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$\sqrt{ \frac{1 + \: sin \: x \: }{1 + \: sin \: x} } = sec \: x \: + \: tan \: x$
$\sqrt{ \frac{1 + sin \: x}{1 + \: sin \: x} } = cosec \: x \: + \: cot \: x \:$
by using trigonometric identities and properties , and correct mathematical properties,
solve L.H.S.
of above statements and prove which statement in true either a) or b).​

Answer 1

Question:

Prove that $\sqrt{\frac{1+\sin x}{1-\sin x}}=\sec x+\tan x$.

Solution:

$\text{LHS}=\sqrt{\frac{1+\sin x}{1-\sin x}}$

Multiply the numerator and denominator by the conjugate $1+\sin x$.

       $=\sqrt{\frac{1+\sin x}{1-\sin x} \times \frac{1+\sin x}{1+\sin x}}$

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

Using the algebraic property, $(a+b)(a-b)=a^2-b^2$ in the denominator

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

We know that $\sin^2x+\cos^2=1 \ \ \ \ \ \ \ \ \Rightarrow(1-\sin^2x)=\cos^2x$

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

Square and square roots are getting cancelled.

      $=\frac{1+\sin x}{\cos x}$

      $=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$

      $=\sec x+\tan x$

      = RHS

LHS = RHS

$\sqrt{\frac{1+\sin x}{1-\sin x}}=\sec x+\tan x$

Hence proved.

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