Question

Question 6
$1+\frac{co{t}^2\alpha }{1+cosec\alpha }=cosec\alpha$

Answer 1

$LHS=1+\frac{co{t}^2\alpha }{1+cosec\alpha }=1+\frac{\frac{co{s}^2\alpha }{si{n}^2\alpha }}{1+\frac{1}{sin\alpha }}$

$[\because cot\theta =\frac{cos\theta }{sin\theta }andcosec\theta =\frac{1}{sin\theta }]$

$=1+\frac{co{s}^2\alpha }{sin\alpha (1+sin\alpha )}=\frac{sin\alpha (1+sin\alpha )+co{s}^2\alpha }{sin\alpha (1+sin\alpha )}$

$=\frac{sin\alpha +(si{n}^2\alpha +co{s}^2\alpha )}{sin\alpha (1+sin\alpha )}[\because si{n}^2\theta +co{s}^2\theta =1]$

$=\frac{(sin\alpha +1)}{sin\alpha (sin\alpha +1)}=\frac{1}{sin\alpha }[\because cosec\theta =\frac{1}{sin\theta }]$

$=cosec\alpha =RHS$