Question

Prove that:
$\frac{sin\theta }{cot\theta +cosec\theta }=2+\frac{sin\theta }{cot\theta -cosec\theta }$

Answer 1

We have to prove that
$\frac{sin\theta }{cot+cosec\theta }=2+\frac{sin\theta }{cot\theta -cosec\theta }or,\frac{sin\theta }{cot\theta +cosec\theta }-\frac{sin\theta }{cot\theta -cosec\theta }=2$

Now,

$\Rightarrow$ LHS = $\frac{sin\theta }{cot\theta +cosec\theta }-\frac{sin\theta }{cot\theta -cosec\theta }$

$\Rightarrow$ LHS = $\frac{sin\theta }{cosec\theta +cot\theta }+\frac{sin\theta }{cosec\theta -cot\theta }$

$\Rightarrow$ LHS = $sin\theta \{\frac{1}{cosec\theta +cot\theta }+\frac{1}{cosec\theta -cot\theta }\}$

$\Rightarrow$ LHS = $sin\theta \left\{\frac{cosec\theta -cot\theta +cosec\theta +cot\theta }{cose{c}^2\theta -co{t}^{}\theta }\right\}=sin\theta \left(\frac{2cosec\theta }{1}\right)$

$\Rightarrow$ LHS = $sin\theta \left(2cosec\theta \right)=2sin\theta \times \frac{1}{sin\theta }=2=RHS$

$\Rightarrow$ LHS = 2 = RHS

ALTERNANATIVELY,

LHS = $\frac{sin\theta }{cot\theta +cosec\theta }$

$\Rightarrow LHS=sin\theta (cosec\theta -cot\theta )$ $[\because \frac{1}{cosec\theta +cot\theta }=cosec\theta -cot\theta ]$

$\Rightarrow$ LHS = $sin\theta (\frac{1}{sin\theta }-\frac{cos\theta }{sin\theta })=sin\theta \left(\frac{1-cos\theta }{sin\theta }\right)$

$\Rightarrow$ LHS = 1 - cos $\theta$

$\Rightarrow$ = 2 - (1 + cos $\theta$)

$\Rightarrow$ LHS = 2 - $\frac{(1+cos\theta )(1-cos\theta )}{1-cos\theta }$

$\Rightarrow$ LHS = 2 - $\frac{(1-co{s}^2\theta )}{1-cos\theta }$

$\Rightarrow$ LHS = 2 - $\frac{si{n}^2\theta }{1-cos\theta }=2-\frac{sin\theta }{\frac{1-cos\theta }{sin\theta }}=2-\frac{sin\theta }{\frac{1}{sin\theta }-\frac{cos\theta }{sin\theta }}$

$\Rightarrow$ LHS = 2 - $\frac{sin\theta }{cosec\theta -cot\theta }$ = RHS