Question

$\bold{if \: \frac{1}{ 1 + \sec \theta} + \frac{1}{ 1 - \sin \theta} = k \sec^{2} \theta } \\ \\ \bold{then \: find \: the \: value \: of \: k \: \: \: \: \: \: \: \: \: \: \: }$

⚠︎ ᴅᴏɴ'ᴛ sᴘᴀᴍ​

Answer 1

Correct Question:-

If $\sf{\dfrac{1}{1+sin\theta} + \dfrac{1}{1-sin\theta} = k sec^2\theta}$, then find the value of k.

Given:-

• $\sf{\dfrac{1}{1+sin\theta} + \dfrac{1}{1-sin\theta} = k sec^2\theta}$

To Find:-

The value of k.

Solution:-

For this question we need to simplify LHS first.

So, firstly taking LHS,

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

Taking LCM,

$\sf{\dfrac{(1-sin\theta) + (1+sin\theta)}{(1+sin\theta)(1-sin\theta)}}$

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

$\sf{\because (a+b)(a-b) = a^2 - b^2}$

= $\sf{\dfrac{1 - \cancel{sin\theta} + 1 + \cancel{sin\theta}}{1-sin^2\theta}}$

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

= $\sf{2\times \dfrac{1}{sec^2\theta}}$

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

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

Since, it is given in the question that,

$\sf{\dfrac{1}{1+sin\theta} + \dfrac{1}{1-sin\theta} = ksec^2\theta}$

Therefore,

$\sf{2sec^2\theta = ksec^2\theta}$

= $\sf{\dfrac{2sec^2\theta}{sec^2\theta} = k}$

= $\sf{\dfrac{2\cancel{sec^2\theta}}{\cancel{sec^2\theta}} = k}$

= $\sf{2 = k}$

=> $\sf{k = 2}$

Therefore, the value of k is 2.

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How did I solve it?

→ For this question we had to solve the LHS first in order to find the value of LHS which is equal to RHS for finding the value of k. So I first solved the LHS and found it's value and after equating LHS with RHS we got the value of k as 2.

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Points to remember!!!

We must know some identities like:-

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

$\sf{\implies sin^2\theta = 1-cos^2\theta}$

$\sf{\implies cos^2\theta = 1-sin^2\theta}$

$\:$

$\sf{1+tan^2\theta = sec^2\theta}$

$\sf{\implies sec^2\theta - tan^\theta = 1}$

$\sf{\implies sec^2\theta - 1 = tan^2\theta}$

$\:$

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

$\sf{\implies cosec^2\theta - cot^2\theta = 1}$

$\sf{\implies cosec^2\theta - 1 = cot^2\theta}$

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Answer 2

sec^2 theta (1+sin thata ) ( 2- sin theta ) = sec ^2 theta ( 1-sin theta) (a+b) (a-b) = (a^2 - b^2) =sec^2 theta .cos ^2 theta =1. ( cos sq. theta + sin sq theta =1) k=1