Question

(sin theta + cos theta )square = 2 ÷ cosec theta × sec theta​

Answer 1

Step-by-step explanation:

We have,

$\rm (\sin θ + \cos θ)^2 = \dfrac{2}{\cosec θ× \sec θ}$

In LHS we have,

$\rm (\sin θ + \cos θ)^2$

By using,

$\boxed{ \rm (a+b)^2= a^2+b^2+2ab}$

$\rm \implies \sin ^2 θ + \cos ^2 θ +2 \sin θ × \cos θ$

We know that,

$\boxed{\rm \sin ^2 θ + \cos ^2 θ =1}$

$\rm \implies 1+ 2 \sin θ × \cos θ$

In RHS we have,

$\rm \dfrac{2}{\cosec θ × \sec θ}$

We know that,

$\boxed{ \: \: \: \cosec θ = \dfrac{1}{\sin θ} \: \: \: }$

and,

$\boxed{ \: \: \: \sec θ = \dfrac{1}{\cos θ} \: \: \: }$

$\rm \implies \dfrac{2}{\dfrac{1}{\sin θ}× \dfrac{1}{\cos θ}}$

$\rm \implies \dfrac{2}{\dfrac{1}{\sin θ× \cos θ}}$

$\rm \implies 2 \sin θ ×\cos θ$

LHS = RHS