Question

Prove that:-

$\frac{ 1+sinA}{1-sinA} = {(secA+tanA)}^{2}$

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Answer 1

Answer is given in the pic.

And don't forget to solve my question.

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\boxed{\bf{Prove\:\:\dfrac{1+\sin A}{1-\sin A}=(\sec A+\tan A)^{2}}}$

♣ ᴛᴏ Pʀᴏvᴇ :

$\sf{\dfrac{1+\sin A}{1-\sin A}=(\sec A+\tan A)^{2}}$

♣ ᴀɴꜱᴡᴇʀ :

$\bf{\dfrac{1+\sin A}{1-\sin A}=(\sec A+\tan A)^{2}}$

Right hand side :

$\bf{(\sec A+\tan A)^{2}=\left(\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\right)^{2}}$

$\sf{=\left(\dfrac{1+\sin A}{\cos A}\right)^{2}}$

$\sf{=\dfrac{(1+\sin A)^{2}}{\cos ^{2} A}}$

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

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

$\sf{=\dfrac{1+\sin A}{1-\sin A}}$

$\boxed{\bf{\bigstar\:\:\dfrac{1+\sin A}{1-\sin A}=(\sec A+\tan A)^{2}}}$

Hence Proved !!!