Question

√1+sinA÷1-sinA + √1-sinA÷1+sinA=2 sec​

Answer 1

the answer is given above

Answer 2

We have to prove that: $\sqrt{\dfrac{1+\sin A}{1-\sin A} } +\sqrt{\dfrac{1-\sin A}{1+\sin A} } =2\sec A$.

L.H.S. = $\sqrt{\dfrac{1+\sin A}{1-\sin A} } +\sqrt{\dfrac{1-\sin A}{1+\sin A} }$

Taking LCM of denominator part, we get

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

Using the algebraic identity:

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

= $\dfrac{1+\sin A+1-\sin A}{\sqrt{1^2-\sin^2 A} }$

= $\dfrac{2}{\sqrt{1-\sin^2 A} }$

Using the trigonometric identity:

$\cos^2 A=1-\sin^2 A$

= $\dfrac{2}{\sqrt{1-\sin^2 A} }$

= $\dfrac{2}{\sqrt{\cos^2 A} }$

= $\dfrac{2}{\cos A} }$

= $2\sec A$

= R.H.S., proved.

Thus, $\sqrt{\dfrac{1+\sin A}{1-\sin A} } +\sqrt{\dfrac{1-\sin A}{1+\sin A} } =2\sec A$, proved.