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Question 1:

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

Solution 1:

1) sinA= 1/cosecA = 1 / √(1+cot²A)

[ cot²A+ 1 = cosec²A,

cosecA= √( 1+cot²A)]

2) tanA= 1/cotA

3) secA= √(1+tan²A)

[sec²A= 1+tan²A ,

secA= √ (1+tan²A)]

secA= √(1+ (1/cot²A)) = √ (1+1/ cot²A)

secA = √(cot²A+1/cot²A)

secA= √1+cot²A/ (cotA)

Comments

Reetika

Question 2:

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

Solution 2:

We know that,

Also, sin² A + cos² A = 1

sin² A = 1 − cos² A

$sinA= \sqrt{1-\left ( \frac{1}{secA} \right )^{2}}$

$=\sqrt{\frac{sec^{2}A-1}{sec^{2}A}}=\frac{\sqrt{sec^{2}A-1}}{secA}$

tan²A + 1 = sec²A

tan²A = sec²A − 1

$tanA=\sqrt{sec^{2}A-1}$

$cotA= \frac{cosA}{sinA}=\frac{\frac{1}{secA}}{\frac{\sqrt{sec^{2}A-1}}{secA}}$

$=\frac{1}{\sqrt{sec^{2}A-1}}$

$cosecA=\frac{1}{sinA}=\frac{secA}{\sqrt{sec^{2}A-1}}$

Comments

Tasneem

Question 3:

Evaluate:

(i) $\frac{sin^{2}63^{\circ}+sin^{2}27^{\circ}}{cos^{2}17^{\circ}+cos^{2}73^{\circ}}$

(ii) $sin25^{\circ}cos65^{\circ}+cos25^{\circ}sin65^{\circ}$

Solution 3:

(i)

sinx= cos(90-x)

sin 63 = cos(90-63) =cos 27

sin 27 = cos(90-27) =cos 63

sin 17 = cos(90-17) =cos 73

sin 73 = cos(90-73)=cos 17

$sin^2 63^{\circ} + sin^2 27^{\circ}}$ = $sin^{2} 63^{\circ} + cos^{2} 63^{\circ}$ = 1

$cos^2 17^{\circ} + cos^{2} 73^{\circ}$ = $cos^{2} 17^{\circ} + sin^{2} 17^{\circ}$ = 1

By using the fact that $sin^{^{2}} + cos^{2} x=1$

Hence, answer is 1

(ii)

sin 25 = cos(90-25) = cos 65

cos 25 = sin(90-25) = sin 65

Hence, cos25 sin65 + cos65 sin25 = cos25 x cos25 + sin25 x sin25 =1

Bu using the fact that $sin^{2}x + cos^{2} x=1$

Comments

Reetika

Question 4:

Choose the correct option. Justify your choice.

(i) $9sec^{2}A-9tan^{2}A=$

(A) 1 (B) 9 (C) 8 (D) 0

(ii) $(1+tan\Theta+sec\Theta)(1+cot\Theta-cosec\Theta )=$

(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) $\frac{1+tan^{2}A}{1+cot^{2}A}$

(A) $sec^{2}A$ (B) $-1$ (C) $cot^{2}A$ (D) $tan^{2}A$

Solution 4:

(i) 9 sec²A − 9 tan²A

= 9 (sec²A − tan²A)

= 9 (1) [As sec² A − tan² A = 1]

= 9

Hence, alternative (B) is correct.

(ii)

(1 + tan θ + sec θ) (1 + cot θ − cosec θ)

$=\left ( 1+\frac{sin\theta }{cos\theta}+\frac{1}{cos\theta } \right )\left ( 1+\frac{cos\theta }{sin\theta }-\frac{1}{sin\theta } \right )$

$=\left ( \frac{cos\theta +sin\theta +1}{cos\theta } \right )$ $\left ( \frac{sin\theta +cos\theta -1}{sin\theta } \right )$

$\frac{\left ( sin\theta +cos\theta \right )^{2}-(1)^{2}}{sin\theta cos\theta }$ $=2$

Hence,option (C) is correct.

(iii) (secA + tanA) (1 − sinA)

$=\left ( \frac{1}{cosA}+\frac{sinA}{cosA} \right )\left ( 1-sinA \right )$

= $\left ( 1+\frac{sinA}{cosA} \right )\left ( 1-sinA \right )$

$=\frac{1-sin^{2}A}{cosA}=\frac{cos^{2}}{cosA}$ = cosA

Hence, option (D) is correct.

(iv) $\frac{1+tan^{2}A}{1+cot^{2}A}$

$\frac{1+\frac{sin^{2}A}{cos^{2}A}}{1+\frac{cos^{2}A}{sin^{2}A}}$ $=\frac{\frac{cos^{2}A+sin^{2}A}{cos^{2}A}}{\frac{sin^{2}A+cos^{2}A}{sin^{2}A}}$

$=\frac{\frac{1}{cos^{2}A}}{\frac{1}{sin^{2}A}}$ $=\frac{sin^{2}A}{cos^{2}A}= tan^{2}A$

Hence, option (D) is correct.

Comments

Tasneem

Question 5:

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) $(cosec\Theta -cot\Theta )^{2}=\frac{1-cos\Theta }{1+cos\Theta }$

(ii) $\frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}=2secA$

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

[Hint : Write the expression in terms of $sin\Theta$ and $cos\Theta$ ]

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

[Hint :Simplify LHS and RHS separately]

(v) $\frac{cosA-sinA+1}{cosA+sinA-1}=cosecA+cotA$ , using the identity $cosec^{2}A=1+cot^{2}A.$

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

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

(viii) $(sinA+cosecA)^{2}+(cosA+secA)^{2}=7+tan^{2}A+cot^{2}A$

(ix) $(cosecA-sinA)(secA-cosA)=\frac{1}{tanA+cotA}$

[Hint: Simplify LHS and RHS separately]

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

Solution 5:

(i)

(ii)

(iii)

= secθ cosec θ +

= R.H.S.

(iv)

= R.H.S

(v)

Using the identity cosec² = 1 + cot²,

L.H.S =

= cosec A + cot A

= R.H.S

(vi)

(vii)

(viii)

(ix)

Hence, L.H.S = R.H.S

(x)

Comments

Tasneem

All Exercises - Chapter 8 - Introduction to Trigonometry

Exercise 8.1

Exercise 8.2

Exercise 8.3

Exercise 8.4

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Learn Exercise 8.4 with Free Lessons & Tips

All Exercises - Chapter 8 - Introduction to Trigonometry

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