Question

sin theta / sin (90- theta) + cos theta/cos (90-theta) = sec theta × cosec theta ​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{\sin\:\theta}{\sin\:(\:90^{\circ}\:-\:\theta\:)}\:+\:\dfrac{\cos\:\theta}{\cos\:(\:90^{\circ}\:-\:\theta\:)}\:=\:\sec\:\theta\:\times\:cosec\:\theta\:}}}$

Step-by-step-explanation:

We have given a trigonometric identity.

We have to prove that identity.

The given trigonometric identity is

$\displaystyle{\sf\:\dfrac{\sin\:\theta}{\sin\:(\:90^{\circ}\:-\:\theta\:)}\:+\:\dfrac{\cos\:\theta}{\cos\:(\:90^{\circ}\:-\:\theta\:)}\:=\:\sec\:\theta\:\times\:cosec\:\theta}$

Now,

$\displaystyle{\sf\:\dfrac{\sin\:\theta}{\sin\:(\:90^{\circ}\:-\:\theta\:)}\:+\:\dfrac{\cos\:\theta}{\cos\:(\:90^{\circ}\:-\:\theta\:)}\:=\:\sec\:\theta\:\times\:cosec\:\theta}$

$\displaystyle{\sf\:LHS\:=\:\dfrac{\sin\:\theta}{\sin\:(\:90^{\circ}\:-\:\theta\:)}\:+\:\dfrac{\cos\:\theta}{\cos\:(\:90^{\circ}\:-\:\theta\:)}}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:\sin\:(\:90^{\circ}\:-\:\theta\:)\:=\:\cos\:\theta\:}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\sin\:\theta}{\cos\:\theta}\:+\:\dfrac{\cos\:\theta}{\cos\:(\:90^{\circ}\:-\:\theta\:)}}$

We know that,

$\displaystyle{\boxed{\blue{\sf\:\cos\:(\:90^{\circ}\:-\:\theta\:)\:=\:\sin\:\theta\:}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\sin\:\theta}{\cos\:\theta}\:+\:\dfrac{\cos\:\theta}{\sin\:\theta\:}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\sin^2\:\theta\:+\:\cos^2\:\theta}{\cos\:\theta\:.\:\sin\:\theta}}$

We know that,

$\displaystyle{\boxed{\green{\sf\:\sin^2\:\theta\:+\:\cos^2\:\theta\:=\:1\:}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{1}{\cos\:\theta\:.\:\sin\:\theta}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{1}{\cos\:\theta}\:\times\:\dfrac{1}{\sin\:\theta}}$

We know that,

$\displaystyle{\boxed{\orange{\sf\:\sec\:\theta\:=\:\dfrac{1}{\cos\:\theta}\:}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\sec\:\theta\:\times\:\dfrac{1}{\sin\:\theta}}$

We know that,

$\displaystyle{\boxed{\purple{\sf\:cosec\:\theta\:=\:\dfrac{1}{\sin\:\theta}\:}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\sec\:\theta\:\times\:cosec\:\theta}$

$\displaystyle{\sf\:RHS\:=\:\sec\:\theta\:\times\:cosec\:\theta}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:LHS\:=\:RHS\:}}}}$

Hence proved!