Question

Plzzz plzz solve it....

URGENT....plzz​

Answer 1

Given:-

$\sf{\sqrt{\dfrac{1+sin\theta}{1-sin\theta}} + \sqrt{\dfrac{1-sin\theta}{1+sin\theta}} = 2 sec\theta}$

To:-

Prove the identity.

Solution:-

Taking LHS,

$\sf{\sqrt{\dfrac{1+sin\theta}{1-sin\theta}} + \sqrt{\dfrac{1-sin\theta}{1+sin\theta}}}$

By Rationalizing the denominator,

$\sf{\sqrt{\dfrac{1+sin\theta}{1-sin\theta}\times \dfrac{1+sin\theta}{1+sin\theta}} + \sqrt{\dfrac{1-sin\theta}{1+sin\theta}\times\dfrac{1-sin\theta}{1-sin\theta}}}$

= $\sf{\sqrt{\dfrac{(1+sin\theta)^2}{1-sin^2\theta}} + \sqrt{\dfrac{(1-sin\theta)^2}{1-sin^2\theta}}}$

= $\sf{\sqrt{\dfrac{(1+sin\theta)^2}{cos^2\theta}}+\sqrt{\dfrac{(1-sin\theta)^2}{cos^2\theta}}}$

= $\sf{\dfrac{1+sin\theta}{cos\theta} + \dfrac{1-sin\theta}{cos\theta}}$

= $\sf{\dfrac{1+sin\theta+1-sin\theta}{cos\theta}}$

= $\sf{\dfrac{2 + \cancel{sin\theta} - \cancel{sin\theta}}{cos\theta}}$

= $\sf{\dfrac{2}{cos\theta}}$

= $\sf{2\times\dfrac{1}{cos\theta}}$

= $\sf{2\times sec\theta}$

= $\sf{2sec\theta}$

Taking RHS,

$\sf{2sec\theta}$

Hence LHS = RHS [Proved]

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Points To Remember!!!

• $\sf{\dfrac{1}{cosec\theta} = sin\theta}$

• $\sf{\dfrac{1}{sin\theta} = cosec\theta}$

• $\sf{\dfrac{1}{sec\theta} = cos\theta}$

• $\sf{\dfrac{1}{cos\theta} = sec\theta}$

• $\sf{\dfrac{1}{cot\theta} = tan\theta}$

• $\sf{\dfrac{1}{tan\theta} = cot\theta}$

• $\sf{\dfrac{sin\theta}{cos\theta} = tan\theta}$

• $\sf{\dfrac{cos\theta}{sin\theta} = cot\theta}$

• $\sf{cos^2\theta + sin^2\theta = 1}$

• $\sf{1+tan^2\theta = sec^2\theta}$

• $\sf{1+cot^2\theta = cosec^2\theta}$

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