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Mathematics · Introduction to Trigonometry · NCERT Exercises

Exercise 8.3

Complete, independently verified solutions for NCERT Exercise 8.3.

Mathematics · Chapter 8 · NCERT Exercises

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Question

Question 1

identities · in terms of cot

Express sin A, sec A and tan A in terms of cot A.

Solution

Board-exam working:

1+cot⁡2A=cosec⁡2A1+\cot^2A=\cosec^2A
cosec⁡A=1+cot⁡2A\cosec A=\sqrt{1+\cot^2A}
sin⁡A=1cosec⁡A=11+cot⁡2A\sin A=\frac1{\cosec A}=\frac1{\sqrt{1+\cot^2A}}
tan⁡A=1cot⁡A\tan A=\frac1{\cot A}
sec⁡2A=1+tan⁡2A=1+1cot⁡2A\sec^2A=1+\tan^2A=1+\frac1{\cot^2A}
sec⁡A=1+cot⁡2Acot⁡A\sec A=\frac{\sqrt{1+\cot^2A}}{\cot A}

Final answer

sin⁡A=11+cot⁡2A,sec⁡A=1+cot⁡2Acot⁡A,tan⁡A=1cot⁡A\sin A=\frac1{\sqrt{1+\cot^2A}},\quad \sec A=\frac{\sqrt{1+\cot^2A}}{\cot A},\quad \tan A=\frac1{\cot A}

Question

Question 2

identities · in terms of sec

Write all other trigonometric ratios of angle A in terms of sec A.

Solution

Board-exam working:

cos⁡A=1sec⁡A\cos A=\frac1{\sec A}
sec⁡2A=1+tan⁡2A⇒tan⁡A=sec⁡2A−1\sec^2A=1+\tan^2A\Rightarrow\tan A=\sqrt{\sec^2A-1}
cot⁡A=1sec⁡2A−1\cot A=\frac1{\sqrt{\sec^2A-1}}
sin⁡A=tan⁡Acos⁡A=sec⁡2A−1sec⁡A\sin A=\tan A\cos A=\frac{\sqrt{\sec^2A-1}}{\sec A}
cosec⁡A=sec⁡Asec⁡2A−1\cosec A=\frac{\sec A}{\sqrt{\sec^2A-1}}

Final answer

cos⁡A=1sec⁡A,tan⁡A=sec⁡2A−1,cot⁡A=1sec⁡2A−1,sin⁡A=sec⁡2A−1sec⁡A,cosec⁡A=sec⁡Asec⁡2A−1\cos A=\frac1{\sec A},\quad \tan A=\sqrt{\sec^2A-1},\quad \cot A=\frac1{\sqrt{\sec^2A-1}},\quad \sin A=\frac{\sqrt{\sec^2A-1}}{\sec A},\quad \cosec A=\frac{\sec A}{\sqrt{\sec^2A-1}}

Question

Question 3

multiple choice · identities

Choose and justify: (i) 9sec²A − 9tan²A; (ii) (1 + tanθ + secθ)(1 + cotθ − cosecθ); (iii) (sec A + tan A)(1 − sin A); (iv) (1 + tan²A)/(1 + cot²A).

Solution

Board-exam working:

(i)

9(sec⁡2A−tan⁡2A)=99(\sec^2A-\tan^2A)=9

Option B.

(ii) Let t = tan θ, so cot θ = 1/t and cosec θ = sec θ/t.

(1+t+sec⁡θ)(1+1t−sec⁡θt)(1+t+\sec\theta)\left(1+\frac1t-\frac{\sec\theta}{t}\right)
=(1+t+sec⁡θ)(1+t−sec⁡θ)t=\frac{(1+t+\sec\theta)(1+t-\sec\theta)}t
=(1+t)2−sec⁡2θt=1+2t+t2−(1+t2)t=2=\frac{(1+t)^2-\sec^2\theta}{t}=\frac{1+2t+t^2-(1+t^2)}t=2

Option C.

(iii)

(sec⁡A+tan⁡A)(1−sin⁡A)=1+sin⁡Acos⁡A(1−sin⁡A)(\sec A+\tan A)(1-\sin A)=\frac{1+\sin A}{\cos A}(1-\sin A)
=1−sin⁡2Acos⁡A=cos⁡A=\frac{1-\sin^2A}{\cos A}=\cos A

Option D.

(iv)

1+tan⁡2A1+cot⁡2A=sec⁡2Acosec⁡2A=tan⁡2A\frac{1+\tan^2A}{1+\cot^2A}=\frac{\sec^2A}{\cosec^2A}=\tan^2A

Option D.

Final answer

(i) B; (ii) C; (iii) D; (iv) D.

Question

Question 4

prove identities · ten subparts

Prove the following identities for acute angles where the expressions are defined:

(i) (cosec⁡θ−cot⁡θ)2=1−cos⁡θ1+cos⁡θ\text{(i) }(\cosec\theta-\cot\theta)^2=\frac{1-\cos\theta}{1+\cos\theta}
(ii) cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2sec⁡A\text{(ii) }\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2\sec A
(iii) tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+sec⁡θcosec⁡θ\text{(iii) }\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}=1+\sec\theta\cosec\theta
(iv) 1+sec⁡Asec⁡A=sin⁡2A1−cos⁡A\text{(iv) }\frac{1+\sec A}{\sec A}=\frac{\sin^2A}{1-\cos A}
(v) cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=cosec⁡A+cot⁡A\text{(v) }\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cosec A+\cot A
(vi) 1+sin⁡A1−sin⁡A=sec⁡A+tan⁡A\text{(vi) }\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A
(vii) sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=tan⁡θ\text{(vii) }\frac{\sin\theta-2\sin^3\theta}{2\cos^3\theta-\cos\theta}=\tan\theta
(viii) (sin⁡A+cosec⁡A)2+(cos⁡A+sec⁡A)2=7+tan⁡2A+cot⁡2A\text{(viii) }(\sin A+\cosec A)^2+(\cos A+\sec A)^2=7+\tan^2A+\cot^2A
(ix) (cosec⁡A−sin⁡A)(sec⁡A−cos⁡A)=1tan⁡A+cot⁡A\text{(ix) }(\cosec A-\sin A)(\sec A-\cos A)=\frac1{\tan A+\cot A}
(x) 1+tan⁡2A1+cot⁡2A=(1−tan⁡A1−cot⁡A)2=tan⁡2A\text{(x) }\frac{1+\tan^2A}{1+\cot^2A}=\left(\frac{1-\tan A}{1-\cot A}\right)^2=\tan^2A

Solution

Board-exam working:

(i)

(cosec⁡θ−cot⁡θ)2=(1−cos⁡θsin⁡θ)2(\cosec\theta-\cot\theta)^2=\left(\frac{1-\cos\theta}{\sin\theta}\right)^2
=(1−cos⁡θ)21−cos⁡2θ=1−cos⁡θ1+cos⁡θ=\frac{(1-\cos\theta)^2}{1-\cos^2\theta}=\frac{1-\cos\theta}{1+\cos\theta}

(ii)

cos⁡A1+sin⁡A+1+sin⁡Acos⁡A\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}
=1−sin⁡Acos⁡A+1+sin⁡Acos⁡A=2cos⁡A=2sec⁡A=\frac{1-\sin A}{\cos A}+\frac{1+\sin A}{\cos A}=\frac2{\cos A}=2\sec A

(iii)

tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}
=sin⁡2θcos⁡θ(sin⁡θ−cos⁡θ)−cos⁡2θsin⁡θ(sin⁡θ−cos⁡θ)=\frac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)}-\frac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)}
=sin⁡3θ−cos⁡3θsin⁡θcos⁡θ(sin⁡θ−cos⁡θ)=\frac{\sin^3\theta-\cos^3\theta}{\sin\theta\cos\theta(\sin\theta-\cos\theta)}
=1+sin⁡θcos⁡θsin⁡θcos⁡θ=1+sec⁡θcosec⁡θ=\frac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta}=1+\sec\theta\cosec\theta

(iv) Left-hand side:

1+sec⁡Asec⁡A=1+cos⁡A\frac{1+\sec A}{\sec A}=1+\cos A

Right-hand side:

sin⁡2A1−cos⁡A=(1−cos⁡A)(1+cos⁡A)1−cos⁡A=1+cos⁡A\frac{\sin^2A}{1-\cos A}=\frac{(1-\cos A)(1+\cos A)}{1-\cos A}=1+\cos A

(v) Use 1 − cos A = sin²A/(1 + cos A).

cos⁡A+sin⁡A−1=sin⁡A−(1−cos⁡A)\cos A+\sin A-1=\sin A-(1-\cos A)
=sin⁡A−sin⁡2A1+cos⁡A=sin⁡A(1+cos⁡A−sin⁡A)1+cos⁡A=\sin A-\frac{\sin^2A}{1+\cos A}=\frac{\sin A(1+\cos A-\sin A)}{1+\cos A}
cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=1+cos⁡Asin⁡A=cosec⁡A+cot⁡A\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\frac{1+\cos A}{\sin A}=\cosec A+\cot A

(vi)

1+sin⁡A1−sin⁡A=(1+sin⁡A)2cos⁡2A\sqrt{\frac{1+\sin A}{1-\sin A}}=\sqrt{\frac{(1+\sin A)^2}{\cos^2A}}
=1+sin⁡Acos⁡A=sec⁡A+tan⁡A=\frac{1+\sin A}{\cos A}=\sec A+\tan A

(vii)

sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=sin⁡θ(1−2sin⁡2θ)cos⁡θ(2cos⁡2θ−1)\frac{\sin\theta-2\sin^3\theta}{2\cos^3\theta-\cos\theta}=\frac{\sin\theta(1-2\sin^2\theta)}{\cos\theta(2\cos^2\theta-1)}
=sin⁡θ(cos⁡2θ−sin⁡2θ)cos⁡θ(cos⁡2θ−sin⁡2θ)=tan⁡θ=\frac{\sin\theta(\cos^2\theta-\sin^2\theta)}{\cos\theta(\cos^2\theta-\sin^2\theta)}=\tan\theta

(viii)

(sin⁡A+cosec⁡A)2+(cos⁡A+sec⁡A)2(\sin A+\cosec A)^2+(\cos A+\sec A)^2
=sin⁡2A+2+cosec⁡2A+cos⁡2A+2+sec⁡2A=\sin^2A+2+\cosec^2A+\cos^2A+2+\sec^2A
=5+(1+cot⁡2A)+(1+tan⁡2A)=7+tan⁡2A+cot⁡2A=5+(1+\cot^2A)+(1+\tan^2A)=7+\tan^2A+\cot^2A

(ix) Left-hand side:

(cosec⁡A−sin⁡A)(sec⁡A−cos⁡A)(\cosec A-\sin A)(\sec A-\cos A)
=1−sin⁡2Asin⁡A⋅1−cos⁡2Acos⁡A=sin⁡Acos⁡A=\frac{1-\sin^2A}{\sin A}\cdot\frac{1-\cos^2A}{\cos A}=\sin A\cos A

Right-hand side:

1tan⁡A+cot⁡A=1sin⁡Acos⁡A+cos⁡Asin⁡A=sin⁡Acos⁡A\frac1{\tan A+\cot A}=\frac1{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}=\sin A\cos A

(x) First expression:

1+tan⁡2A1+cot⁡2A=sec⁡2Acosec⁡2A=tan⁡2A\frac{1+\tan^2A}{1+\cot^2A}=\frac{\sec^2A}{\cosec^2A}=\tan^2A

Second expression:

(1−tan⁡A1−cot⁡A)2=(1−tan⁡A1−1tan⁡A)2\left(\frac{1-\tan A}{1-\cot A}\right)^2=\left(\frac{1-\tan A}{1-\frac1{\tan A}}\right)^2
=(1−tan⁡Atan⁡A−1tan⁡A)2=(−tan⁡A)2=tan⁡2A=\left(\frac{1-\tan A}{\frac{\tan A-1}{\tan A}}\right)^2=(-\tan A)^2=\tan^2A

Final answer

All ten identities are proved.