Question

(ix) cot-¹[√1-sinx + √1+ sinx /
√1- sinx -
√1+ sinx ]​

Answer 1

Answer:

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Answer 2

Answer:

π - x/2

Step-by-step explanation:

The question is $cot^{-1} (\frac{\sqrt{1-sinx}+\sqrt{1+sinx} }{\sqrt{1-sinx}-\sqrt{1+sinx} } )$

Lets give a substitution for$\sqrt{1 +sinx\\}$

We know,

$cos^{2}x + sin^{2} x = 1$

$sin2x = 2sinx cosx$

So,

1 + sinx = $cos^{2}x + sin^{2} x + 2sin\frac{x}{2}cos\frac{x}{2} = (cos\frac{x}{2} +sin\frac{x}{2} )^{2}$

$\sqrt{1+sinx} = (cos\frac{x}{2}+sin\frac{x}{2} )$ ____(1)

Similarly, $\sqrt{1-sinx}$ = $(cos\frac{x}{2} - sin\frac{x}{2})$ ____(2)

Substituting in the question,

$cot^{-1} (\frac{\sqrt{1-sinx}+\sqrt{1+sinx} }{\sqrt{1-sinx}-\sqrt{1+sinx} } )$

= $cot^{-1} ( \frac{ (cos\frac{x}{2}-sin\frac{x}{2} )+ (scos\frac{x}{2}+sin\frac{x}{2} )}{\ (cos\frac{x}{2}-sin\frac{x}{2} )- (cos\frac{x}{2}+sin\frac{x}{2} )})$

= $cot^{-1} ( \frac{ cos\frac{x}{2}-sin\frac{x}{2} + cos\frac{x}{2}+sin\frac{x}{2} }{\ cos\frac{x}{2}-sin\frac{x}{2} - cos\frac{x}{2}-sin\frac{x}{2} })$

= $cot^{-1} ( \frac{2cos\frac{x}{2} }{-2sin\frac{x}{2} } )$

= $cot^{-1} ( - cot\frac{x}{2} )$

We know,

$cot^{-1} (-x) = Pi - cot^{-1}x$

So, $cot^{-1} ( - cot\frac{x}{2} ) = Pi - cot^{-1}(cot\frac{x}{2} ) = Pi - \frac{x}{2}$

So, Answer is Pi - x/2