NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4

NCERT Solutions Class 10 Maths
Chapter – 8 (Introduction to Trigonometry)

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 8 Introduction to Trigonometry Exercise 8.4 has been provided here to help the students in solving the questions from this exercise.

Chapter : 8 Introduction to Trigonometry

Exercise – 8.4

1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Solution –
cosec²A– cot²A = 1
⇒ cosec²A = 1 + cot²A
⇒ 1/sin²A = 1 + cot²A
⇒ sin²A = 1/(1+cot²A)
⇒ sin A = $\frac{\pm 1}{\sqrt{1+cot^2A}}$

Now,
sin²A = 1/(1+cot²A)
⇒ 1 – cos²A = $\frac{1}{(1+cot^2A)}$
⇒ cos²A = 1 – $\frac{1}{(1+cot^2A)}$
⇒ cos²A = (1 + cot²A -1)/(1 + cot²A)
⇒ 1/sec²A = cot²A/(1+cot²A)
⇒ sec A = (1 + cot²A)/cot²A
⇒ sec A = $\frac{\pm \sqrt{1+cot^2A}}{cotA}$

also,
tan A = sin A/cos A
cot A = cos A/sin A
⇒ tan A = 1/cot A

2. Write all the other trigonometric ratios of ∠A in terms of sec A.

Solution – We know that,
sec A = 1/cos A
⇒ cos A = 1/sec A
also,
cos²A + sin²A = 1
⇒ sin²A = 1 – cos²A
⇒ sin²A = 1 – (1/sec²A)
⇒ sin²A = (sec²A – 1)/sec²A
⇒ sin A = $\frac{\pm \sqrt{sec^2A -1}}{secA}$

also,
sin A = 1/cosec A
⇒ cosec A = 1/sin A
⇒ cosec A = $\frac{\pm secA}{\sqrt{sec^2A-1}}$

Now,
sec²A – tan²A = 1
⇒ tan²A = sec²A + 1
⇒ tan A = $\sqrt{sec^2A +1}$

also,
tan A = 1/cot A
⇒ cot A = 1/tan A
⇒ cot A = $\frac{\pm1}{\sqrt{sec^2A +1}}$

3. Evaluate:
(i) (sin²63° + sin²27°)/(cos²17° + cos²73°)
(ii) sin 25° cos 65° + cos 25° sin 65°

Solution –

(i) (sin²63° + sin²27°)/(cos²17° + cos²73°)
= (sin²(90° – 27°) + sin²27°) / (cos²(90° – 73°) + cos²73°))
= (cos²27°+ sin²27°)/(sin²27° + cos²73°)
= 1/1 (∵ sin²A + cos²A = 1)
= 1

(ii) sin 25° cos 65° + cos 25° sin 65°
= sin(90° – 25°) cos 65° + cos(90° – 65°) sin 65°
= cos 65° cos 65° + sin 65° sin 65°
= cos²65°+ sin²65°
= 1

4. Choose the correct option. Justify your choice.
(i) 9 sec²A – 9 tan²A =
(A) 1 (B) 9 (C) 8 (D) 0

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
(A) 0 (B) 1 (C) 2 (D) – 1

(iii) (sec A + tan A) (1 – sin A) =
(A) sec A (B) sin A (C) cosec A (D) cos A

(iv) 1+tan²A/1+cot²A =
(A) sec²A (B) -1 (C) cot²A (D) tan²A

Solution –

(i) 9 sec²A – 9 tan²A
= 9 (sec²A – tan²A)
= 9 × 1 (∵ sec2 A – tan2 A = 1)
= 9
(B) is correct.

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
= (1 + sin θ/cos θ + 1/cos θ) (1 + cos θ/sin θ – 1/sin θ)
= (cos θ + sin θ + 1)/cos θ × (sin θ + cos θ – 1)/sin θ
= (cos θ + sin θ)²– 1²/(cos θ sin θ)
= (cos²θ + sin²θ + 2cos θ sin θ – 1)/(cos θ sin θ)
= (1 + 2cos θ sin θ – 1)/(cos θ sin θ)
= (2cos θ sin θ)/(cos θ sin θ)
= 2
(C) is correct

(iii) (secA + tanA) (1 – sinA)
= (1/cos A + sin A/cos A) (1 – sinA)
= (1 + sin A/cos A) (1 – sinA)
= (1 – sin²A)/cos A
= cos²A/cos A
= cos A
(D) is correct.

(iv) (1 + tan²A)/(1 + cot²A)
= (1 + 1/cot²A)/(1 + cot²A)
= (cot²A + 1/cot²A) /(1 + cot²A)
= 1/cot²A
= tan²A
(D) is correct.

5. Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.
(i) (cosec θ – cot θ)²= (1-cos θ)/(1+cos θ)
(ii) $\frac{coaA}{1+sinA} + \frac{1+sinA}{cosA}$ = 2 sec A

(iii) $\frac{tan\theta}{1-cot\theta} +\frac{cot\theta}{1-tan\theta}$ = 1 + sec θ cosec θ [Hint : Write the expression in terms of sin θ and cos θ]
(iv) (1 + sec A)/sec A = sin²A/(1-cos A) [Hint : Simplify LHS and RHS separately]

(v) $\frac{cosA - sin A + 1}{cos A+sinA-1}$ = cosec A + cot A, using the identity cosec²A = 1 + cot²A.

(vi) $\sqrt{\frac{1+sinA}{1-sinA}} = secA + tanA$

(vii) $\frac{(sin\theta-2sin^3\theta)}{2cos^3\theta-cos\theta}$ = tan θ

(viii) (sin A + cosec A)²+ (cos A + sec A)² = 7 + tan²A + cot²A

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A + cotA) [Hint : Simplify LHS and RHS separately]

(x) $\left ( \frac{1+tan^2A}{1+cot^2A} \right ) = \left ( \frac{1-tanA}{1-cotA} \right )^2 = tan^2A$

Solution –

(i) (cosec θ – cot θ)²= (1 – cos θ)/(1 + cos θ)
L.H.S. = (cosec θ – cot θ)²= (cosec²θ + cot²θ – 2cosec θ cot θ)
= (1/sin²θ + cos²θ/sin²θ – 2cos θ/sin²θ)
= (1 + cos²θ – 2cos θ)/(1 – cos²θ)
= (1 – cos θ)²/(1 – cosθ)(1 + cos θ)
= (1 – cos θ)/(1 + cos θ)
= R.H.S.

(ii) $\frac{coaA}{1+sinA} + \frac{1+sinA}{cosA}$ = 2 sec A
L.H.S. = $\frac{coaA}{1+sinA} + \frac{1+sinA}{cosA}$
= [cos²A + (1 + sin A)²]/(1 + sin A) cos A
= (cos²A + sin²A + 1 + 2sin A)/(1 + sin A) cos A
Since cos²A + sin²A = 1
= (1 + 1 + 2sin A)/(1 + sin A) cos A
= (2+ 2sin A)/(1 + sin A) cos A
= 2(1 + sin A)/(1 + sin A)cos A
= 2/cos A
= 2 sec A
= R.H.S.

(iii) $\frac{tan\theta}{1-cot\theta} +\frac{cot\theta}{1-tan\theta}$ = 1 + sec θ cosec θ

L.H.S. = $\frac{tan\theta}{1-cot\theta} +\frac{cot\theta}{1-tan\theta}$

We know that tan θ = sin θ/cos θ
cot θ = cos θ/sin θ
= [(sin θ/cos θ)/1 – (cos θ/sin θ)] + [(cos θ/sin θ)/1 – (sin θ/cos θ)]
= [(sin θ/cos θ)/(sin θ – cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ – sin θ)/cos θ]
= sin²θ/[cos θ(sin θ – cos θ)] + cos²θ/[sin θ(cos θ – sin θ)]
= sin²θ/[cos θ(sin θ – cos θ)] – cos²θ/[sin θ(sin θ – cos θ)]
= 1/(sin θ – cos θ) [(sin²θ/cos θ) – (cos²θ/sin θ)]
= 1/(sin θ – cos θ) × [(sin³θ – cos³θ)/sin θ cos θ]
= [(sin θ – cos θ)(sin²θ + cos²θ + sin θ cos θ)]/[(sin θ – cos θ)sin θ cos θ]
= (1 + sin θ cos θ)/sin θ cos θ
= 1/sin θ cos θ + 1
= 1 + sec θ cosec θ
= R.H.S.

(iv) $\frac{1+secA}{secA} = \frac{sin^2A}{1-cosA}$

First find the simplified form of L.H.S
L.H.S. = $\frac{1+secA}{secA}$
= $\frac{1+\frac{1}{cosA}}{\frac{1}{cosA}}$

= $\frac{\frac{cosA+1}{cosA}}{\frac{1}{cosA}}$
= cos A + 1

R.H.S. = sin²A/(1-cos A)
= $\frac{(1-cos^2A)}{1-cosA}$

= (1 – cos A)(1 + cos A)/(1 – cos A)
= cos A + 1
L.H.S. = R.H.S.
Hence proved

(v) $\frac{cosA - sin A + 1}{cos A+sinA-1}$ = cosec A + cot A

L.H.S. = $\frac{cosA - sin A + 1}{cos A+sinA-1}$

= $\frac{\frac{cosA - sin A + 1}{sinA}}{\frac{cos A+sinA-1}{sinA}}$
We know that cos A/sin A = cot A and 1/sin A = cosec A
= (cot A – 1 + cosec A)/(cot A + 1 – cosec A)
= (cot A – cosec²A + cot²A + cosec A)/(cot A + 1 – cosec A) (using cosec²A – cot²A = 1)
= ((cot A + cosec A) – (cosec²A – cot²A))/(cot A + 1 – cosec A)
= ((cot A + cosec A) – (cosec A + cot A)(cosec A – cot A))/(1 – cosec A + cot A)
= (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)
= cot A + cosec A
= R.H.S.
Hence Proved

(vi) $\sqrt{\frac{1+sinA}{1-sinA}} = secA + tanA$

L.H.S. = $\sqrt{\frac{1+sinA}{1-sinA}}$

= $\sqrt{\frac{\frac{1+sinA}{cosA}}{\frac{1-sinA}{cosA}}}$
We know that 1/cos A = sec A and sin A/cos A = tan A
= $\sqrt{\frac{secA+tanA}{secA-tanA}}$
Now using rationalization, we get

= $\sqrt{\frac{secA+tanA}{secA-tanA}}$ × $\sqrt{\frac{secA + tanA}{secA + tanA}}$

= $\sqrt{\frac{(secA+tanA)^2}{sec^2A-tan^2A}}$

= (sec A + tan A)/1
= sec A + tan A
= R.H.S
Hence proved

(vii) $\frac{(sin\theta-2sin^3\theta)}{2cos^3\theta-cos\theta}$ = tan θ

L.H.S. = $\frac{(sin\theta-2sin^3\theta)}{2cos^3\theta-cos\theta}$
= (sin θ (1 – 2sin²θ))/(cos θ(2cos²θ – 1))
We know that sin²θ = 1 – cos²θ
= (sin θ(1 – 2(1 – cos²θ))/(cos θ(2cos²θ – 1))
= (sin θ(2cos²θ – 1))/(cos θ(2cos²θ -1))
= sin θ/ cos θ
= tan θ
= R.H.S.
Hence proved

(viii) (sin A + cosec A)²+ (cos A + sec A)² = 7+tan²A+cot²A
L.H.S. = (sin A + cosec A)²+ (cos A + sec A)²(a+b)² = a² + b² +2ab
= (sin²A + cosec²A + 2 sin A cosec A) + (cos²A + sec²A + 2 cos A sec A)
= (sin²A + cos²A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan²A + 1 + cot²A
= 1 + 2 + 2 + 2 + tan²A + cot²A
= 7 + tan²A + cot²A
= R.H.S.
Hence proved.

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
L.H.S. = (cosec A – sin A)(sec A – cos A)
= (1/sin A – sin A)(1/cos A – cos A)
= ((1 – sin²A)/sin A][(1 – cos²A)/cos A)
= (cos²A/sin A) × (sin²A/cos A)
= cos A sin A

R.H.S. = 1/(tan A + cotA)
= 1/(sin A/cos A + cos A/sin A)
= 1/[(sin²A + cos²A)/sin A cos A]
= cos A sin A

L.H.S. = R.H.S.
Hence proved

(x) $\left ( \frac{1+tan^2A}{1+cot^2A} \right ) = \left ( \frac{1-tanA}{1-cotA} \right )^2 = tan^2A$

L.H.S. = $\left ( \frac{1+tan^2A}{1+cot^2A} \right )$
= (1 + tan²A)/(1 + 1/tan²A))
= (1 + tan²A)/((1 + tan²A)/tan²A)
= tan²A

Similarly,

$\left ( \frac{1-tanA}{1-cotA} \right )^2$ =tan²A

Hence proved

Go Back To Chapters

NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4