Question

If sinA=(m-1)/(m+1), then prove that, tanA*cosecA =(m+1)/2√m​

Answer 1

Answer:

Given:

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

To prove:

$\sf{tan \ A.cosec \ A=\dfrac{m+1}{2\sqrt{m}}}$

Solution:

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

$\sf{\therefore{cos^{2}A=1-(\dfrac{m-1}{m+1})^{2}}}$

$\sf{\therefore{cos^{2}A=1-\dfrac{(m-1)^{2}}{(m+1)^{2}}}}$

$\sf{\therefore{cos^{2}A=\dfrac{(m+1)^{2}-(m-1)^{2}}{(m+1)^{2}}}}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\sf{\therefore{cos \ A=\dfrac{4\sqrt{m}}{m+1}}}$

$\sf{\underline{\underline{Proof:}}}$

$\sf{L.H.S.= tan \ A.cosec \ A }$

$\sf{[But \ cosec\theta=\dfrac{1}{sin\theta}]}$

$\sf{=tan \ A\times\dfrac{1}{sin \ A}}$

$\sf{[But \ tan\theta=\dfrac{sin\theta}{cos\theta}]}$

$\sf{=\dfrac{sin \ A}{cos A}\times\dfrac{1}{sin \ A}}$

$\sf{=\dfrac{1}{cos \ A}}$

$\sf{=\dfrac{1}{\frac{2\sqrt{m}}{m+1}}}$

$\sf{=\dfrac{m+1}{2\sqrt{m}}}$

$\sf{=R.H.S.}$

$\sf{Hence, \ proved.}$