NCERT Solutions for Class 10 Maths Chapter 8 - Introduction to Trigonometry
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Exercise 8.1
1. In ΔABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine
(i) sin A, cos A
(ii) sin C, cos C
Sol. (i) In right angled ΔABC, by using the Pythagoras theorem, we have
AC2 = AB2 + BC2
= (24)2 + (7)2
= 576 + 49 = 625
∴ AC = 25 cm
(∵ Side cannot be –ve)
$$\text{Now, sin A =}\frac{\text{P}}{\text{H}}=\frac{\text{BC}}{\text{AC}}=\frac{\text{7}}{\text{25}}\\\text{and}\space\text{cos A}=\frac{\text{B}}{\text{H}}=\frac{\text{AB}}{\text{AC}}=\frac{\text{24}}{\text{25}}\\\text{(ii)}\space\text{sin C}=\frac{\text{P}}{\text{H}}=\frac{\text{AB}}{\text{AC}}=\frac{24}{25}\\\text{and}\space\text{cos C}=\frac{\text{B}}{\text{H}}=\frac{\text{CB}}{\text{AC}}=\frac{7}{25}$$
2. In figure, find tan P – cot R.
Sol. In right angled ΔPQR, By Pythagoras theorem
PQ2 + QR2 = PR2
⇒ (12)2 + QR2 = (13)2
⇒ 144 + QR2 = 169
⇒ QR2 = 169 – 144 = 25
∴ QR = 5 cm
(∵ Side cannot be negative)
$$\text{tan}\space\text{P}=\frac{\text{P}}{\text{B}}=\frac{\text{QR}}{\text{PQ}}=\frac{5}{12}\\\text{and \space cot R=}\frac{\text{B}}{\text{P}}=\frac{\text{QR}}{\text{PQ}}=\frac{5}{12}\\\therefore\space\text{tan P – cot R =}\frac{5}{12}-\frac{5}{12}=0$$
$$\textbf{3. If sin A =}\frac{3}{4}\space,\textbf{calculate cos A and tan A.}$$
$$\textbf{Sol.}\space\text{Given,\space sin A =}\frac{3}{4}\\\Rarr\space\frac{\text{P}}{\text{H}}=\frac{\text{3}}{\text{4}}$$
Let BC = P = 3k and AC = H = 4k.
In right angled ΔABC,
P2 + B2 = H2
(By Pythagoras theorem)
⇒ (3k)2 + AB2 = (4k)2
⇒ 9k2 + AB2 = 16k2
⇒ AB2 = 16k2 – 9k2 = 7k2
$$\therefore\space\text{AB}+k\sqrt{7}\\(\because\space\text{Base cannot be – ve})\\\text{Now, cosA}=\frac{\text{B}}{\text{H}}=\frac{\text{AB}}{\text{AC}}\\=\frac{k\sqrt{7}}{4k}=\frac{\sqrt{7}}{4}\\\text{and\space tanA =}\frac{\text{P}}{\text{B}}=\frac{\text{BC}}{\text{AB}}=\frac{3k}{k\sqrt{7}}\\=\frac{3}{\sqrt{7}}×\frac{\sqrt{7}}{7}=\frac{3\sqrt{7}}{7}$$
4. Given 15 cot A = 8, find sin A and sec A.
Sol. Given, 15 cot A = 8
$$\Rarr\space\text{cot A}=\frac{8}{15}\\\Rarr\space\frac{\text{B}}{\text{P}}=\frac{8}{15}=\frac{\text{AB}}{\text{BC}}$$
Let AB = B = 8k and BC = P = 15k
In right angled ΔABC, by pythagoras theorem
H2 = B2 + P2
AC2 = AB2 + BC2
= (8k)2 + (15k)2
= 64k2 + 225k2
= 289k2
∴ AC = 17k (∵ Side cannot be – ve)
$$\text{Now, sin A =}\frac{\text{P}}{\text{H}}=\frac{\text{BC}}{\text{AC}}\\=\frac{\text{15k}}{\text{17k}}=\frac{\text{15}}{\text{17}}\\\text{and sec A =}\frac{\text{H}}{\text{B}}=\frac{\text{AC}}{\text{AB}}\\=\frac{\text{17 k}}{\text{8k}}=\frac{17}{8}$$
$$\textbf{5. Given sec}\space\theta =\frac{13}{12},\textbf{calculate all other trigonometric ratios.}$$
$$\textbf{Sol.}\space\text{Given \space sec}\space\theta=\frac{13}{12}\\\Rarr\space\frac{\text{H}}{\text{B}}=\frac{13}{12}$$
Let H = 13k, B = 12k
In right angled triangle, by pythagoras theorem
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ P2 + (12k)2 = (13k)2
⇒ P2 + 144k2 = 169k2
⇒ P2 = 169k2 – 144k2 = 25k2
⇒ P = 5k (∵ Side cannot be –ve)
$$\text{Now, \space sin}\space\theta=\frac{\text{P}}{\text{H}}=\frac{5k}{13k}=\frac{5}{13}\\\text{cos}\space\theta=\frac{\text{B}}{\text{H}}=\frac{12k}{13k}=\frac{12}{13}\\\text{tan\space}\theta=\frac{\text{P}}{\text{B}}=\frac{5k}{12k}=\frac{5}{12}\\\text{cosec}\space\theta =\frac{1}{\text{sin}\theta}=\frac{13}{5}\\\text{and\space cot}\theta=\frac{1}{\text{tan}\space\theta}=\frac{12}{5}$$
6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Sol. Given : ΔABC, in which
cos A = cos B
To prove : ∠A = ∠B
Proof : Now, cos A = cos B
$$\therefore\space\frac{\text{AC}}{\text{AB}}=\frac{\text{BC}}{\text{AB}}\\\bigg(\because\space\text{cos}\space\theta=\frac{\text{Base}}{\text{Hypotenunse}}\bigg)$$
⇒ AC = BC
⇒ ∠A = ∠B
(∵ Angle opposite to equal sides are also equal)
Hence Proved.
$$\textbf{7. If cot} \theta =\frac{\textbf{7}}{\textbf{8}},\textbf{evaluate}\\\textbf{(i)}\space\frac{(\textbf{1}+\textbf{sin}\theta)(1-\textbf{sin}\theta)}{(\textbf{1}+\textbf{cos}\theta)(\textbf{1}-\textbf{cos}\theta)}$$
(ii) cot2 θ
$$\textbf{Sol.}\space\text{Given, \space cot}\space\theta =\frac{7}{8}\\\Rarr\space\frac{\text{B}}{\text{P}}=\frac{\text{7}}{\text{8}}$$
Let B = 7k, P = 8k
In a right angled ΔABC, by pythagoras theorem
H2 = B2 + P2
= 49k2 + 64k2
= 113k2
$$\therefore\space\text{H}=k\sqrt{113}\\(\because\space\text{Side cannot be –ve})\\\text{Now, sin} \theta=\frac{\text{P}}{\text{H}}=\frac{\text{8k}}{k\sqrt{113}}=\frac{\text{8}}{\sqrt{113}}\\\text{and \space cos}\space\theta=\frac{\text{B}}{\text{H}}=\frac{7k}{k\sqrt{113}}=\frac{7}{\sqrt{113}}\\\text{(i)}\space\frac{(1+\text{sin}\theta)(1-\text{sin}\theta)}{(1+\text{cos}\theta)(1-\text{cos}\theta)}\\=\frac{1^{2}-\text{sin}^{2}\theta}{1^{2}-\text{cos}^{2}\theta}\\\lbrack\because(a + b)(a – b) = a^2 – b^2\rbrack\\=\frac{1-\bigg(\frac{8}{\sqrt{113}}\bigg)^{2}}{1-\bigg(\frac{7}{\sqrt{113}}\bigg)^{2}}$$
$$=\frac{1-\frac{64}{113}}{1-\frac{49}{113}}=\frac{\frac{113-64}{113}}{\frac{113-49}{113}}=\frac{49}{64}\\\textbf{(ii)\space}\text{cot}^{2}\theta=\bigg(\frac{7}{8}\bigg)^{2}=\frac{49}{64}\\\bigg(\because\space\text{Given cot}\theta=\frac{7}{8}\bigg)$$
8. If 3 cot A = 4, check whether
$$\frac{\textbf{1}-\textbf{tan}^{2}\textbf{A}}{\textbf{1}+\textbf{tan}^{2}\textbf{A}}=\textbf{cos}^{\textbf{2}}\textbf{A\space– sin}^\textbf{2} \textbf{A or not.}$$
Sol. Given, 3 cot A = 4
$$\therefore\space\text{cot A}=\frac{4}{3}\\\Rarr\space\frac{\text{B}}{\text{P}}=\frac{4}{3}=\frac{\text{AB}}{\text{BC}}$$
In right angled ΔABC, by pythagoras theorem
AC2 = AB2 + BC2
H2 = P2 + B2
= (3k)2 + (4k)2
= 9k2 + 16k2 = 25k2
∴ AC = H = 5k (∵ Side cannot be –ve)
$$\text{Now, tan A}=\frac{1}{\text{cot}A}=\frac{3}{4}\space\bigg(\because\text{cotA}=\frac{4}{3}\bigg)\\\text{cos A}=\frac{\text{B}}{\text{H}}=\frac{\text{AB}}{\text{AC}}=\frac{4}{5}\\\text{and\space sin A}=\frac{\text{P}}{\text{H}}=\frac{\text{BC}}{\text{AC}}=\frac{3}{5}\\\text{LHS}=\frac{1-\text{tan}^{2}\text{A}}{1+\text{tan}^{2}\text{A}}\\=\frac{1-\bigg(\frac{3}{4}\bigg)^{2}}{1+\bigg(\frac{3}{4}\bigg)^{2}}\\=\frac{\frac{16-9}{16}}{\frac{16+9}{16}}=\frac{7}{25}$$
...(i)
and RHS = cos2 A – sin2 A
$$=\bigg(\frac{4}{5}\bigg)^{2}-\bigg(\frac{3}{5}\bigg)^{2}\\=\frac{16-9}{25}=\frac{7}{25}\space\text{...(ii)}$$
From equations (i) and (ii), w
e get
LHS = RHS
$$\textbf{9. In triangle ABC, right angled at B, if tan A =}\frac{\textbf{1}}{\sqrt{\textbf{3}}},$$
find the value of
(ii) sin A cos C + cos A sin C
(iii) cos A cos C – sin A sin C
$$\textbf{Sol.}\space\text{Given,\space tan A =}\frac{1}{\sqrt{3}}\\\Rarr\space\frac{\text{P}}{\text{B}}=\frac{1}{\sqrt{3}}\\\text{Let P = k, B =}\sqrt{3k}$$
In a right angled by pythagoras theorem
B2 + P2 = H2
$$\Rarr\space(\sqrt{3}k)^{2}+(1k)^{2}=\text{H}^{2}$$
⇒ 3k2 + k2 = H2
⇒ H2 = 4k2
∴ H = 2k (∵ Side cannot be –ve)
(i) sin A cos C + cos A sin C
$$=\bigg(\frac{\text{P}}{\text{H}}\bigg)\bigg(\frac{\text{B}}{\text{H}}\bigg)+\bigg(\frac{\text{B}}{\text{H}}\bigg)\bigg(\frac{\text{P}}{\text{H}}\bigg)\\\bigg(\frac{k}{2k}\bigg)\bigg(\frac{k}{2k}\bigg)+\bigg(\frac{k\sqrt{3}}{2k}\bigg)\bigg(\frac{k\sqrt{3}}{2k}\bigg)\\=\frac{k^{2}}{4k^{2}}+\frac{3k^{2}}{4k^{2}}=\frac{4k^{2}}{4k^{2}}=\frac{4}{4}$$
= 1
(ii) cos A cos C – sin A sin C
$$=\bigg(\frac{\text{P}}{\text{H}}\bigg)\bigg(\frac{\text{P}}{\text{H}}\bigg)-\bigg(\frac{\text{B}}{\text{H}}\bigg)\bigg(\frac{\text{B}}{\text{H}}\bigg)\\=\bigg(\frac{k}{2k}\bigg)\bigg(\frac{k\sqrt{3}}{2k}\bigg)-\bigg(\frac{k\sqrt{3}}{2}\bigg)\bigg(\frac{k}{2k}\bigg)\\=\frac{1}{2}×\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}×\frac{1}{2}\\=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0$$
10. In ΔPQR, right angled at Q, PR +QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Sol. Let QR = x and PR = y
Given, PR + QR = 25
⇒ y + x = 25
⇒ y = (25 – x) ...(i)
11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
$$\textbf{(ii) sec A =}\frac{\textbf{12}}{\textbf{5}}\space\textbf{for some value of angle A.}$$
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
$$\textbf{(v)sin}\space\theta=\frac{\textbf{4}}{\textbf{3}}\textbf{for some angle}\space\theta.$$
Sol. (i) False, because the value of tan A increases from 0 to ∞. Also, tan 45° = 1.
(ii) True, because the value of sec A increase from 1 to ∞.
(iii) False, because cos A is the abbreviation used for the cosine of angle A. And cosecant k is for cosec A.
(iv) False, because cot A is one symbol. We cannot seperate cot and A.
(v) False, because the value of sin θ always lies between 0 and 1.
$$\text{Here, sin}\space\theta =\frac{4}{3}$$
which is greater than 1. So, it is not possible.
Exercise 8.2
1. Evaluate the following
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan2 45° + cos2 30° – sin2 60°
$$\textbf{(iii)}\space\frac{\textbf{cos 45}\degree}{\textbf{sec 30}\degree+\textbf{cosec 30}\degree}\\\textbf{(iv)\space}\frac{\textbf{sin 30}\degree+\textbf{tan 45}\degree-\textbf{cosec 60}\degree}{\textbf{sec 30}\degree+\textbf{cos 30}\degree+\textbf{cos 45}\degree}\\\textbf{(v)\space}\frac{\textbf{5}\space\textbf{cos}^{2}\textbf{60}\degree+\textbf{4}\space\textbf{sec}^{2}\textbf{30}\degree-\textbf{tan}^{2}\textbf{45}\degree}{\textbf{sin}^{2}\textbf{30}\degree+\textbf{cos}^{2}\textbf{30}\degree}$$
Sol. (i) sin 60° cos 30° + sin 30° cos 60°
On putting the values
$$=\frac{\sqrt{3}}{2}×\frac{\sqrt{3}}{2}+\frac{1}{2}×\frac{1}{2}\\=\frac{3}{4}+\frac{1}{4}\\=\frac{3+1}{4}\\=\frac{4}{4}=1$$
(ii) 2 tan2 45° + cos2 30° – sin2 60°
On putting the values
$$2(1)^{2}+\bigg(\frac{\sqrt{3}}{2}\bigg)^{2}-\bigg(\frac{\sqrt{3}}{2}\bigg)^{2}\\=2(1)+\frac{3}{4}-\frac{3}{4}=2$$
$$\textbf{(iii)\space}\frac{\text{cos 45\degree}}{\text{sec 30\degree + cosec 30\degree}}$$
On putting the values
$$=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + \frac{2}{1}}\\=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2\sqrt{3}}{\sqrt{3}}}=\frac{1}{\sqrt{2}}×\frac{\sqrt{3}}{(2+2\sqrt{3})}\\=\frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}×\frac{2\sqrt{2}-2\sqrt{6}}{2\sqrt{2}-2\sqrt{6}}\\=\frac{2\sqrt{6}-2\sqrt{18}}{(2\sqrt{2})^{2}-2\sqrt{6})^{2}}\\=\frac{2\sqrt{6}-2\sqrt{18}}{(2\sqrt{2})^{2}-(2\sqrt{6})^{2}}\\=\frac{2\sqrt{6}-2(3\sqrt{2})}{8-24}$$
$$=\frac{-2(3\sqrt{2})+2\sqrt{6}}{-16}=\frac{-2(3\sqrt{2}-\sqrt{6})}{-16}\\=\frac{3\sqrt{2}-\sqrt{6}}{8}$$
$$\textbf{(iv)}\space\frac{\text{sin 30}\degree+\text{tan 45}\degree-\text{cosec 60}\degree}{\text{sec 30}\degree+\text{cos 60}\degree+\text{cot 45}\degree}$$
On putting the values we get
$$=\frac{\frac{1}{2}+\frac{1}{1}-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}\\=\frac{\frac{\sqrt{3}+2\sqrt{3}-4}{2\sqrt{3}}}{\frac{4+\sqrt{3}+2\sqrt{3}}{2\sqrt{3}}}\\=\frac{3\sqrt{3}-4}{4+3\sqrt{3}}×\frac{4-3\sqrt{3}}{4-3\sqrt{3}}$$
$$\text{(Multiply by conjugate of }4+3\sqrt{3}\\\space\text{in numerator and denominator both)}\\=\frac{12\sqrt{3}-27-16+12\sqrt{3}}{16-27}\\=\frac{24\sqrt{3}-43}{-11}\\\frac{-(43-24\sqrt{3})}{11}\\=\frac{43-24\sqrt{3}}{11}$$
$$\textbf{(v)}\space\frac{5\text{cos}^{2}60\degree+4\text{sec}^{2}30\degree-\text{tan}^{2}45\degree}{\text{sin}^{2}30\degree+\text{cos}^{2}30\degree}$$
On putting the values, we get
$$=\frac{5\bigg(\frac{1}{2}\bigg)^{2}+4\bigg(\frac{2}{\sqrt{3}}\bigg)^{2}-(1)^{2}}{\bigg(\frac{1}{2}\bigg)^{2}+\bigg(\frac{\sqrt{3}}{2}\bigg)^{2}}\\=\frac{5\bigg(\frac{1}{4}\bigg)+4\bigg(\frac{4}{3}\bigg)-1}{\frac{1}{4}+\frac{3}{4}}\\=\frac{\frac{(15+64-12)}{12}}{\bigg(\frac{1+3}{4}\bigg)}=\frac{67}{12}×\frac{4}{4}=\frac{67}{12}$$
2. Choose the correct option and justify your choice
$$\textbf{(i)}\space\frac{2\space\textbf{tan}\textbf{30}\degree}{1+\textbf{tan}^{2}\textbf{30}\degree}\textbf{=}$$
(a) sin 60°
(b) cos 60°
(c) tan 60°
(d) sin 30°
$$\textbf{(ii)}\space\frac{1-\textbf{tan}^{\textbf{2}}\textbf{45}\degree}{1+\textbf{tan}^{2}\textbf{45}\degree}=$$
(a) tan 90°
(b) 1
(c) sin 45°
(d) 0
(iii) sin 2A = 2 sin A is true when A =
(a) 0°
(b) 30°
(c) 45°
(d) 60°
$$\textbf{(iv)}\space\frac{2\textbf{tan}\textbf{30}\degree}{1-\textbf{tan}^{2}\textbf{30}\degree}\textbf{=}$$
(a) cos 60°
(b) sin 60°
(c) tan 60°
(d) sin 30°
Sol. (i) (a) sin 60°
Explanation :
$$\frac{2 \textbf{tan} \textbf{30}\degree}{\textbf{1+ tan}^{2}\textbf{30}\degree}=\frac{2\bigg(\frac{1}{\sqrt{3}}\bigg)}{1+\bigg(\frac{1}{\sqrt{3}}\bigg)^{2}}=\frac{\frac{2}{\sqrt{3}}}{\bigg(1+\frac{1}{3}\bigg)}\\=\frac{\frac{2}{\sqrt{3}}}{\bigg(\frac{3+1}{3}\bigg)}=\frac{2}{\sqrt{3}}×\frac{3}{4}\\\frac{\sqrt{3}}{2}=\text{sin}\space 60\degree$$
(ii) (d) 0
Explanation :
$$\frac{1-\text{tan}^{2}45\degree}{1+\text{tan}^{2}45\degree}=\frac{1-(1)^{2}}{1+(1)^{2}}\\=\frac{1-1}{1+1}=\frac{0}{2}=0$$
(iii) (a) 0°
Explanation :
Given, sin 2A = 2 sin A
When A = 0°, sin (2 × 0°) = 2 sin (0°)
⇒ sin(0°) = 2(0)
0 = 0, True
(iv) (c) tan 60°
Explanation :
$$\frac{2\text{tan}30\degree}{1-\text{tan}^{2}30\degree}=\frac{2\bigg(\frac{1}{\sqrt{3}}\bigg)}{1-\bigg(\frac{1}{\sqrt{3}}\bigg)^{2}}\\=\frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}}\\\frac{2}{\sqrt{3}}×\frac{3}{2}=\frac{3}{\sqrt{3}}\\=\sqrt{3}=\text{tan 60°}$$
$$\textbf{3. If tan (A + B) =} \sqrt{3}\space\textbf{and tan(A – B) =}\frac{\textbf{1}}{\sqrt{\textbf{3}}};\\\textbf{0° < A + B ≤ 90°; A < B, find A and B.}$$
$$\textbf{Sol.}\space\text{We have, tan(A + B) =}\sqrt{3}$$
⇒ tan(A + B) = tan 60°
∴ A + B = 60° ...(i)
$$\text{and tan(A – B) =}\frac{1}{\sqrt{3}}$$
⇒ tan(A – B) = tan 30°
∴ A – B = 30° ...(ii)
On adding (i) and (ii), we get
⇒ A = 45°
⇒ B = 60° – 45°
⇒ B = 15°
Hence, A = 45° and B = 15°.
4. State whether the following are true or false. Justify your answer.
(i) sin(A + B) = sin A + sin B
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ
(v) cot A is not defined for A = 0°.
Sol. (i) False; because
When A = 60° and B = 30°
sin(A + B) = sin(60° + 30°)
= sin 90° = 1
and sin A + sin B = sin 60° + sin 30°
$$=\frac{\sqrt{3}}{2}+\frac{1}{2}\\=\frac{\sqrt{3}+1}{2}$$
So, sin(A + B) ≠ sin A + sin B
(ii) True; because, it is clear from the table below that the sin θ increase as θ increase.
| θ | sin θ |
| 0° | 0 |
| 30° | $$\frac{1}{2}$$ |
| 45° | $$\frac{1}{\sqrt{2}}$$ |
| 60° | $$\frac{\sqrt{3}}{2}$$ |
| 90° | 1 |
(iii) False; because it is clear from the table below that the cos θ does not increase as θ increase and it decreases.
| θ | cos θ |
| 0° | 1 |
| 30° | $$\frac{\sqrt{3}}{2}$$ |
| 45° | $$\frac{1}{\sqrt{2}}$$ |
| 60° | $$\frac{1}{2}$$ |
| 90° | 0 |
(iv) False; because it is true only for θ = 45°
$$\bigg(\because\text{sin}45\degree=\frac{1}{\sqrt{2}}=\text{cos}45\degree\bigg)$$
(v) True; since for A = 0°, tan A = tan 0° = 0
$$\therefore\space\text{cot A}=\frac{1}{\text{tan A}}=\frac{1}{0}=\text{not defined (∞)}$$
Exercise 8.3
1. Evaluate
$$\textbf{(i)}\space\frac{\textbf{sin}\textbf{18}\degree}{\textbf{cos}\textbf{72}\degree}\\\textbf{(ii)}\space \frac{\textbf{tan 26}\degree}{\textbf{cot 64}\degree}$$
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°
$$\textbf{Sol. \space(i)}\space\frac{\text{sin}18\degree}{\text{cos}72\degree}=\frac{\text{sin}(90\degree-72\degree)}{\text{cos}\space 72\degree}\\=\frac{\text{cos}\space72\degree}{\text{cos}\space72\degree}=1\\\lbrack\because\space\text{sin}(90\degree-\theta)=\text{cos}\theta\rbrack\\\textbf{(ii)\space}\frac{\text{tan}\space26\degree}{\text{cot}\space64\degree}\\=\frac{\text{tan}26\degree}{\text{cot}(90\degree-26\degree)}\\\frac{\text{tan 26}\degree}{\text{tan 26}\degree}$$
[∵ cot(90° – θ) = tan θ]
= 1
(iii) cos 48° – sin 42° = cos(90° – 42°) – sin 42°
= sin 42° – sin 42°
[∵ cos(90° – θ) = sin θ]
= 0
(iv) cosec 31° – sec 59°
= cosec 31° – sec(90° – 31°)
= cosec 31° – cosec 31°
[∵ sec(90° – θ) = cosec θ]
= 0
2. Show that
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Sol. (i) LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48°. tan 23° tan(90° – 48°). tan(90° – 23°)
= tan 48°. tan 23°. cot 48°. cot 23°
[∵ tan(90° – θ) = cot θ]
$$=\text{tan 48°. tan 23° .}\frac{1}{\text{tan 48}\degree}.\frac{1}{\text{tan 23}\degree}\\\bigg(\because\text{cot}\theta=\frac{1}{\text{tan}\space\theta}\bigg)$$
= 1 = R.H.S
(ii) LHS = cos 38° cos 52° – sin 38° sin 52°
= cos(90° – 52°) cos(90° – 38°)– sin 38° sin 52°
= sin 52° sin 38° – sin 38° sin 52°
[∵ cos(90° – θ) = sin θ]
= 0 = RHS
3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Sol. Given, tan 2A = cot(A – 18°)
⇒ cot(90° – 2A) = cot(A – 18°)
[∵ tan θ = cot(90° – θ)]
On comparing the angles on both sides, we get
⇒ 90° – 2A = A – 18°
⇒ 108° = 3A
$$\Rarr\space\text{A}=\frac{108\degree}{3}=36\degree$$
4. If tan A = cot B, prove that A + B = 90°
Sol. Given, tan A = cot B
⇒ cot(90° – A) = cot B
[∵ tan θ = cot (90° – θ)]
On comparing the angles on both sides, we get
⇒ 90° – A = B
∴ 90° = A + B
or A + B = 90°
5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Sol. Given, sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A – 20°)
[∵ sec θ = cosec (90° – θ)]
On comparing the angles on both sides, we get
⇒ 90° – 4A = A – 20°
⇒ 90° + 20° = A + 4A
⇒ 110° = 5A
$$\Rarr\space\text{A}=\frac{110\degree}{5}=22\degree$$
6. If A, B and C are interior angles of a triangle ABC, then show that
$$\textbf{sin}\bigg(\frac{\textbf{B+C}}{\textbf{2}}\bigg)=\textbf{cos}\frac{\textbf{A}}{\textbf{2}}$$
Sol. We know that,
Sum of all angles in a triangle is 180°.
∴ A + B + C = 180°
$$\Rarr\space\frac{\text{A}}{2}+\frac{\text{B}}{2}+\frac{\text{C}}{2}=90\degree\\\Rarr\space\frac{\text{B}}{2}+\frac{\text{C}}{2}=90\degree-\frac{\text{A}}{2}$$
Taking sine of angle on both sides, we get
$$\text{sin}\bigg(\frac{\text{B+C}}{2}\bigg)=\text{sin}\bigg(90\degree-\frac{\text{A}}{2}\bigg)\\\Rarr\space\text{sin}\bigg(\frac{\text{B+C}}{2}\bigg)=\text{cos}\frac{\text{A}}{2}$$
[∵ sin (90° – θ) = cos θ]
Hence Proved.
7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Sol. We have,
sin 67° + cos 75° = sin(90° – 23°) + cos(90° – 15°)
= cos 23° + sin 15°
[∵ sin(90° – θ) = cos θ and cos(90° – θ) = sin θ]
Exercise 8.4
1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Sol. (i) By identity,
cosec2 A = 1 + cot2 A
$$\Rarr\space\text{cosec A}=\sqrt{1+\text{cot}^{2}\text{A}}\\\text{Now, sin A =}\frac{1}{\text{cosec A}}\\=\frac{1}{\sqrt{1+\text{cot}^{2}\text{A}}}\\\bigg(\because\space\text{sin A}=\frac{1}{\text{cosec A}}\bigg)\\\textbf{(ii)}\space\text{tan A =}\frac{1}{\text{cot A}}\\\bigg(\because\text{tan A}=\frac{1}{\text{cot A}}\bigg)$$
(iii) By identity,
sec2 A = 1 + tan2 A
$$\Rarr\space\text{sec A =}\sqrt{1+\text{tan}^{2} \text{A}}\\=\sqrt{1+\bigg(\frac{1}{\text{cot A}}\bigg)^{2}}\\=\sqrt{1+\frac{1}{\text{cot}^{2}\text{A}}}\\\bigg(\because\text{tan A}=\frac{1}{\text{cot A}}\bigg)\\=\sqrt{\frac{1+\text{cot}^{2}\text{A}}{\text{cot}^{2}\text{A}}}\\=\frac{\sqrt{1+\text{cot}^{2}\text{A}}}{\text{cot A}}$$
2. Write all the other trigonometric ratios of ∠A in terms of sec A.
Sol. (i) By identity,
sin2 A + cos2 A = 1
⇒ sin2 A = 1 – cos2 A
$$\Rarr\space\text{sin A =}\sqrt{1-\text{cos}^{2}\text{A}}\\=\sqrt{1-\frac{1}{\text{sec}^{2}\text{A}}}\\\bigg(\because\text{cos A}=\frac{1}{\text{sec A}}\bigg)\\=\sqrt{\frac{\text{sec}^{2}\text{A}-1}{\text{sec}^{2}A}}\\=\frac{\sqrt{\text{sec}^{2}\text{A}-1}}{\text{sec A}}\\\textbf{(ii)\space}\text{cos A =}\frac{1}{\text{sec A}}\space\bigg(\because\text{cos A}=\frac{1}{\text{sec A}}\bigg)$$
(iii) We know that,
sec2 A = 1 + tan2 A
⇒ tan2 A = sec2 A – 1
$$\Rarr\space\text{tan A}=\sqrt{\text{sec}^{2}\text{A}-1}\\\textbf{(iv)}\space\text{cot A}=\frac{1}{\text{tan A}}=\frac{1}{\sqrt{sec^{2}\text{A-1}}}\\\text{[By using part (iii)]}\\\textbf{(v)\space}\text{cosec A =}\frac{1}{\text{sin A}}=\frac{\text{sec A}}{\sqrt{\text{sec}^{2}}\text{A}-1}$$
[By using part (i)]
3. Evaluate
$$\textbf{(i)\space}\frac{\textbf{sin}^{2}\textbf{63}\degree+\textbf{sin}^{2}\textbf{27}\degree}{\textbf{cos}^{2}\textbf{17}\degree+\textbf{cos}^{2}\textbf{73}\degree}$$
(ii) sin 25° cos 65° + cos 25° sin 65°
$$\textbf{Sol. (i)}\space\frac{\text{sin}^{2}63\degree+\text{sin}^{2}27\degree}{\text{cos}^{2}17\degree+\text{cos}^{2}73\degree}\\=\frac{\text{sin}^{2}63\degree+\text{sin}^{2}(90\degree-63\degree)\degree}{\text{cos}^{2}17\degree+\text{cos}^{2}(90\degree-17\degree)}\\\frac{\text{sin}^{2}63\degree+\text{cos}^{2}63\degree}{\text{cos}^{2}17\degree+\text{sin}^{2}17\degree}\\\begin{bmatrix}\because\space\text{sin(90\degree-}\theta)=\text{cos}\theta\\\text{and cos(90\degree-}\theta)=\text{sin}\theta\end{bmatrix}\\=\frac{1}{1}=1\\(\because \space\text{sin}^{2}\theta+\text{cos}^{2}\theta=1)$$
(ii) sin 25° cos 65° + cos 25° sin 65°
= sin 25° cos (90° – 25°) + cos 25° sin (90° – 25°)
= sin 25° sin 25° + cos 25° cos 25°
$$\begin{bmatrix}\because\space\text{cos(90\degree-}\theta)=\text{sin}\theta\\\text{and \space sin(90\degree-}\theta)=\text{cos}\theta\end{bmatrix}$$
= sin2 25° + cos2 25° (∵ sin2 θ + cos2 θ = 1)
= 1.
5. Prove the following identities, where the angle involved are acute angles for which the expressions are defined.
$$\textbf{(i)\space (cosec} \theta – \textbf{cot} \theta)2=\frac{1-\textbf{cos}\theta}{1+\textbf{cos}\theta}\\\textbf{(ii)\space}\frac{\textbf{cos A}}{1+\textbf{sin A}}+\frac{\textbf{1+sin A}}{\textbf{cos A}}=2\space\textbf{sec A}\\\textbf{(iii)\space}\frac{\textbf{tan}\theta}{1-\textbf{cot}\theta}+\frac{\textbf{cot}\theta}{1-\textbf{tan}\theta}=1+\textbf{sec}\theta\space\textbf{cosec}\theta$$
[Hint : Write the expression in terms of sin θ and cos θ]
$$\textbf{(iv)\space}\frac{\textbf{1}+\textbf{sec A}}{\textbf{sec A}}=\frac{\textbf{sin}^{2}\textbf{A}}{\textbf{1- cos A}}\\\textbf{(v)\space}\frac{\textbf{cos A}-\textbf{sin A}+\textbf{1}}{\textbf{cos A}+\textbf{sin A}-\textbf{1}}=$$
= cosec A + cot A
Using the identity
cosec2 A = 1 + cot2 A
$$\textbf{(vi)}\space\sqrt{\frac{\textbf{1+sin A}}{\textbf{1-sin A}}}\\\textbf{= sec A + tan A}\\\textbf{(vii)\space}\frac{\textbf{sin}\theta-2\textbf{sin}^{3}\theta}{2\textbf{cos}^{3}\theta-\textbf{cos}\theta}$$
= tan θ
(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
$$\textbf{(ix) (cosec A – sin A)(sec A – cos A) =}\frac{1}{\textbf{tan A + cot A}}$$
[Hint : Simplify LHS and RHS separately]
$$\textbf{(x)}\space\bigg(\frac{\textbf{1+tan}^{2}\textbf{A}}{\textbf{1+cot}^{2}\textbf{A}}\bigg)=\bigg(\frac{\textbf{1-tan A}}{\textbf{1-cot A}}\bigg)^{2}=\textbf{tan}^{2}\textbf{A}$$
Sol. (i) LHS = (cosec θ – cot θ)2
$$=\bigg(\frac{1}{\text{sin}\theta}-\frac{\text{cos}\theta}{\text{sin}\theta}\bigg)^{2}=\bigg(\frac{\text{1-cos}\theta}{\text{sin}\theta}\bigg)^{2}\\=\frac{(1-\text{cos}\theta)^{2}}{\text{sin}^{2}\theta}=\bigg(\frac{(1-\text{cos}\theta)^{2}}{1-\text{cos}^{2}\theta}\bigg)$$
(∵ sin2 θ + cos2 θ = 1)
$$=\frac{(1-\text{cos}\theta)(1-\text{cos}\theta)}{(1+\text{cos}\theta)(1-\text{cos}\theta)}$$
[∵ a2 – b2 = (a + b)(a – b)]
$$=\frac{1-\text{cos}\theta}{1+\text{cos}\theta}=\text{R.H.S}$$
Hence Proved.
$$\textbf{(ii)\space}\text{L.H.S}=\frac{\text{cosA}}{1+\text{sin A}}+\frac{1+\text{sin A}}{\text{cos A}}\\=\frac{\text{cos}^{2}\text{A}+(1+\text{sin A})^{2}}{(1+\text{sin A}).\text{cos A}}\\=\frac{\text{cos}^{2}\text{A}+\text{sin}^{2}\text{A}+1+2\text{sin A}}{(1+\text{sin A}).\text{cos A}}\\\lbrack\because\space(a+b)^{2}=a^{2}+b^{2}+2ab\rbrack\\=\frac{1+1+2\text{sin A}}{(1+\text{sin A}).\text{cos A}}$$
(∵ sin2 A + cos2 A = 1)
$$=\frac{2+2\space\text{sin A}}{(1+\text{sin A}).\text{cos A}}\\=\frac{2(1+\text{sin A})}{(1+ \text{sin A}).\text{cos A}}\\=\frac{2}{\text{cos A}}$$
= 2 sec A = R.H.S.
$$\bigg[\because\space\frac{1}{\text{cos A}}=\text{sec A}\bigg]$$
Hence Proved.
$$\text{(iii)\space LHS}=\frac{\text{tan}\theta}{\text{1-cot}\theta}+\frac{cot\theta}{\text{1-\text{tan}}\theta}\\=\frac{\frac{\text{sin}\theta}{\text{cos}\theta}}{1-\frac{\text{cos}\theta}{\text{sin}\theta}}+\frac{\frac{\text{cos}\theta}{\text{sin}\theta}}{1-\frac{\text{sin}\theta}{\text{cos}\theta}}\\\begin{pmatrix}\because\text{tan}\theta=\frac{\text{sin}\theta}{\text{cos}\theta}\\\text{and cot}\theta=\frac{\text{cos}\theta}{\text{sin}\theta}\end{pmatrix}\\=\frac{\frac{\text{sin}\theta}{\text{cos}\theta}}{\frac{\text{sin}\theta-\text{cos}\theta}{\text{sin}\theta}}+\frac{\frac{\text{cos}\theta}{\text{sin}\theta}}{\frac{\text{cos}\theta-\text{sin}\theta}{\text{cos}\theta}}$$
$$=\frac{\text{sin}\theta}{\text{cos}\theta}×\frac{\text{sin}\theta}{\text{sin}\theta-\text{cos}\theta}+\frac{\text{cos}\theta}{\text{sin}\theta}×\frac{\text{cos}\theta}{\text{cos}\theta-\text{sin}\theta}\\=\frac{\text{sin}^{2}\theta}{\text{cos}\theta(\text{sin}\theta-\text{cos}\theta)}-\frac{\text{cos}^{2}\theta}{\text{sin}\theta(\text{sin}\theta-\text{cos}\theta)}\\=\frac{\text{sin}^{3}\theta-\text{cos}^{3}\theta}{\text{cos}\theta.\text{sin}\theta(\text{sin}\theta-\text{cos}\theta)}\\=\frac{(\text{sin}\theta-\text{cos}\theta).(\text{sin}^{2}\theta+\text{cos}^{2}\theta+\text{sin}\theta cos\theta)}{\text{cos}\theta.\text{sin}\theta.(\text{sin}\theta-\text{cos}\theta)}$$
[∵ a3 – b3 = (a – b)(a2 + b2 + ab)]
$$=\frac{1+\text{sin}\theta\text{cos}\theta}{\text{cos}\theta\text{sin}\theta}\\(\because\space\text{sin}^{2}\theta+\text{cos}^{2}\theta=1)\\=\frac{1}{\text{cos}\theta.\text{sin}\theta}+\frac{\text{sin}\theta.\text{cos}\theta}{\text{sin}\theta.\text{cos}\theta}\\=\frac{1}{\text{cos}\theta.\text{sin}\theta}+1$$
= 1 + sec θ . cosec θ = RHS
Hence Proved.
$$\textbf{(iv)}\space\text{LHS =}\frac{1+\text{secA}}{\text{secA}}=\frac{1+\frac{1}{\text{cos A}}}{\frac{1}{\text{cos A}}}\\\bigg(\because\text{secA}=\frac{1}{\text{cos A}}\bigg)\\=\frac{\frac{\text{cos A+1}}{\text{cos A}}}{\frac{1}{\text{cos A}}}\\\frac{\text{cos A+1}}{\text{cos A}}×\frac{\text{cos A}}{1}=\text{cos A + 1}$$
LHS = 1 + cos A ...(i)
$$\text{Now, RHS =}\frac{\text{sin}^{2}\text{A}}{\text{1-cos A}}=\frac{1-\text{cos}^{2}\text{A}}{\text{1-cos A}}$$
(∵ sin2 A + cos2 A = 1)
$$=\frac{(1+\text{cos A})(1-\text{cos A})}{(1-\text{cos A})}\\=\lbrack\because\space a^{2}-b^{2}=(a+b)(a-b)\rbrack$$
= 1 + cos A ...(ii)
From equations (i) and (ii), we get
LHS
= RHS = 1 Hence Proved.
$$\textbf{(v)}\space\text{LHS}=\frac{\text{cos A - sin A + 1}}{\text{cos A + sin A - 1}}$$
On dividing numerator and denominator by sin A, we get
$$\text{LHS = }\frac{\frac{\text{cos A}}{\text{sin A}}-\frac{\text{sin A}}{\text{sin A}}+\frac{\text{1}}{\text{sin A}}}{\frac{\text{cos A}}{\text{sin A}}+\frac{\text{sin A}}{\text{sin A}}-\frac{\text{1}}{\text{sin A}}}\\\qquad=\frac{\text{cot A-1 + cosec A}}{\text{cot A+1 - cosec A}}\\=\frac{\text{cot A + cosec A- 1}}{\text{cot A+1 - cosec A}}\\=\frac{(\text{cot A + cosec A})-(\text{cosec}^{2}\text{A}-\text{cot}^{2}\text{A})}{\text{cot A + 1 - cosec A}}$$
(∵ 1 = cosec2 A – cot2 A)
$$\frac{(\text{cot A + cosec A})-[(\text{cosec A + cot A}) (\text{cosec A - cot A})]}{\text{cot A + 1 - cosec A}}$$
[∵ (a2 – b2) = (a + b)(a – b)]
$$=\frac{(\text{cot A + cosec A})[1-(\text{cosec A - cot A})]}{\text{cot A + 1 - cosec A}}$$
$$=\frac{(\text{cot A + cosec A})[1-\text{cosec A + cot A}]}{\text{cot A + 1 - cosec A}}$$
= cot A + cosec A = RHS
Hence Proved.
$$\textbf{(vi)\space}\text{LHS}=\sqrt{\frac{1+\text{sin A}}{1-\text{sin A}}}\\=\sqrt{\frac{1+\text{sin A}}{1-\text{sin A}}×\frac{1+\text{sin A}}{1+\text{sin A}}}\\\text{(on rationalisation)}\\=\sqrt{\frac{(1+\text{sin A})^{2}}{1-\text{sin}^{2}A}}\\=\sqrt{\frac{(1+\text{sin A})^{2}}{\text{cos}^{2}A}}\\(\because\space\text{sin}^{2}A + \text{cos}^{2}A=1)\\=\frac{1+\text{sin A}}{\text{cos A}}\\=\frac{1}{\text{cos A}}+\frac{\text{sin A}}{\text{cos A}}$$
= sec A + tan A
$$\bigg[\because\frac{1}{\text{cos A}}=\text{sec A}\space\text{and}\space\frac{\text{sin A}}{\text{cos A}}=\text{tan A}\bigg]$$
= RHS
Hence Proved.
$$\textbf{(vii)}\space\text{LHS}=\frac{\text{sin}\theta-2\text{sin}^{3}\theta}{2\text{cos}^{3}\theta-\text{cos}\theta}\\=\frac{\text{sin}\theta(1-2\text{sin}^{2}\theta)}{\text{cos}\theta(2 \text{cos}^{2}\theta-1)}\\=\frac{\text{sin}\theta[1-2(1-\text{cos}^{2}\theta)]}{\text{cos}\theta(2\text{cos}^{2}\theta-1)}$$
(∵ sin2 θ+ cos2θ = 1)
$$=\frac{\text{sin}\theta[1-2+2\text{cos}^{2}\theta]}{\text{cos}\theta[2\text{cos}^{2}\theta-1]}\\\frac{\text{sin}\theta(2\text{cos}^{2}\theta-1)}{\text{cos}\theta(2\space\text{cos}^{2}\theta-1)}\\=\frac{\text{sin}\space\theta}{\text{cos}\space\theta}$$
= tan θ =RHS Hence Proved.
(viii) LHS = (sin A + cosec A)2 + (cos A + sec A)2
= sin2 A + cosec2 A + 2 sin A cosec A + cos2 A + sec2 A + 2 cos A sec A
[∵ (a + b)2 = a2 + b2 + 2ab]
= (sin2 A + cos2 A) + (1 + cot2 A)
$$+\text{2 sin A}.\frac{1}{\text{sin A}}+(1+\text{tan}^{2}\text{A})\\+ 2\text{cos A}.\frac{1}{\text{cos A}}$$
[∵ sin2 A + cos2 A = 1, 1 + cot2 A = cosec2 A, 1 + tan2 A = sec2 A)
= 1 + 1 + cot2 A + 2 + 1 + tan2 A + 2
= 7 + tan2 A + cot2 A
= RHS Hence Proved.
(ix) LHS = (cosec A – sin A)(sec A – cos A)
$$=\bigg(\frac{1}{\text{sin A}}-\frac{\text{sin A}}{1}\bigg)\bigg(\frac{1}{\text{cos A}}-\frac{\text{cos A}}{1}\bigg)\\\bigg[\because\space\text{cosec A}=\frac{1}{\text{sin A}},\text{sec A}=\frac{1}{\text{cos A}}\bigg]\\=\bigg(\frac{1-\text{sin}^{2}\text{A}}{\text{sin A}}\bigg)\bigg(\frac{1-\text{cos}^{2}\text{A}}{\text{cos A}}\bigg)\\=\frac{\text{cos}^{2}\text{A}}{\text{sin A}}×\frac{\text{sin}^{2}\text{A}}{\text{cos A}}$$
(∵ sin2 A + cos2 A = 1)
= cos A sin A ...(i)
Now,
$$\text{RHS}=\frac{1}{\text{tan A + cot A}}\\=\frac{1}{\bigg(\frac{\text{sin A}}{\text{cos A}}+\frac{\text{cos A}}{\text{sin A}}\bigg)}\\=\frac{1}{\bigg(\frac{\text{sin}^{2}\text{A}+\text{cos}^{2}\text{A}}{\text{cos A sin A}}\bigg)}=\frac{1}{\frac{1}{\text{cos A sin A}}}$$
(∵ sin2 A + cos2 A = 1)
$$1×\frac{\text{cos A}\text{sin A}}{1}$$
= cos A sin A ...(ii)
From equations (i) and (ii), we get
LHS = RHS
$$\textbf{(x)\space}\text{LHS}=\bigg(\frac{1+\text{tan}^{2}\text{A}}{1+\text{cot}^{2}\text{A}}\bigg)\\=\frac{\frac{1}{1}+\frac{\text{sin}^{2}\text{A}}{\text{cos}^{2}\text{A}}}{\frac{1}{1}+\frac{\text{cos}^{2}\text{A}}{\text{sin}^{2}\text{A}}}\\\begin{bmatrix}\because\space\text{tan A}=\frac{\text{sin A}}{\text{cos A}}\\\text{and cotA =}\frac{\text{cos A}}{\text{sin A}}\end{bmatrix}\\=\frac{\frac{\text{cos}^{2}\text{A}+\text{sin}^{2}\text{A}}{\text{cos}^{2}\text{A}}}{\frac{\text{cos}^{2}\text{A}+\text{cos}^{2}\text{A}}{\text{cos}^{2}\text{A}}}=\frac{\frac{1}{\text{cos}^{2}\text{A}}}{\frac{1}{\text{sin}^{2}\text{A}}}$$
(∵ sin2 A + cos2 A = 1)
$$=\frac{1}{\text{cos}^{2}\text{A}}×\frac{\text{sin}^{2}\text{A}}{1}\\=\frac{\text{sin}^{2}\text{A}}{\text{cos}^{2}\text{A}}=\text{tan}^{2}\text{A}\\= \text{RHS}$
Second part :
$$\bigg(\frac{\text{1-\text{tan A}}}{1-\text{cot A}}\bigg)^{2}=\bigg(\frac{1-\frac{\text{sin A}}{\text{cos A}}}{1-\frac{\text{cos A}}{\text{sin A}}}\bigg)^{2}$$
$$\begin{bmatrix}\because\space\text{tan A}=\frac{\text{sin A}}{\text{cos A}}\\\text{and cot A}=\frac{\text{cos A}}{\text{sin A}}\end{bmatrix}\\=\begin{pmatrix}\frac{\frac{\text{cos A-sin A}}{\text{cos A}}}{\frac{\text{sin A-cos A}}{\text{sin A}}}\end{pmatrix}^{2}\\=\bigg(\frac{\text{cos A-sin A}}{\text{cos A}}×\frac{\text{sin A}}{\text{sin A - cos A}}\bigg)^{2}\\=\bigg(-1×\frac{\text{sin A}}{\text{cos A}}\bigg)^{2}$$
= (– tan A)2 = tan2 A = RHS
Hence Proved.
Science Most Likely Question Bank
CBSE Class 10 for 2025 Exam
