Question

Prove the following identities : (i) sin 0 cot 0 + sin 0 cosec 0 = 1 + cos O​

Answer 1

Step-by-step explanation:

sin 0 cot 0 = sin 0 x cos 0/ sin 0 = cos 0

sin 0 cosec 0 = sin 0 x 1/sin 0 = 1

So, sin 0 cot 0 + sin 0 cosec 0 = cos 0 + 1

Answer 2

Answer:

$\displaystyle{\boxed{\red{\sf\:\sin\:\theta\:\cot\:\theta\:+\:\sin\:\theta\:cosec\:\theta\:=\:1\:+\:\cos\:\theta\:}}}$

Step-by-step-explanation:

We have given a trigonometric identity.

We have to prove that identity.

The given trigonometric identity is

$\displaystyle{\sf\:\sin\:\theta\:\cot\:\theta\:+\:\sin\:\theta\:cosec\:\theta\:=\:1\:+\:\cos\:\theta}$

$\displaystyle{\sf\:LHS\:=\:\sin\:\theta\:\cot\:\theta\:+\:\sin\:\theta\:cosec\:\theta}$

$\displaystyle{\implies\sf\:LHS\:=\:\sin\:\theta\:(\:\cot\:\theta\:+\:cosec\:\theta\:)}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:\cot\:\theta\:=\:\dfrac{1}{\tan\:\theta}\:}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\sin\:\theta\:\left(\:\dfrac{1}{\tan\:\theta}\:+\:cosec\:\theta\:\right)}$

We know that,

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

$\displaystyle{\implies\sf\:LHS\:=\:\sin\:\theta\:\left(\:\dfrac{1}{\tan\:\theta}\:+\:\dfrac{1}{\sin\:\theta}\:\right)}$

$\displaystyle{\implies\sf\:LHS\:=\:\sin\:\theta\:\left(\:\dfrac{\sin\:\theta\:+\:\tan\:\theta}{\tan\:\theta\:.\:\sin\:\theta}\:\right)}$

We know that,

$\displaystyle{\boxed{\green{\sf\:\tan\:\theta\:=\:\dfrac{\sin\:\theta}{\cos\:\theta}\:}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\sin\:\theta\:\left(\:\dfrac{\sin\:\theta\:+\:\dfrac{\sin\:\theta}{\cos\:\theta}}{\dfrac{\sin\:\theta}{\cos\:\theta}\:\times\:\sin\:\theta}\:\right)}$

$\displaystyle{\implies\sf\:LHS\:=\:\sin\:\theta\:\left(\:\dfrac{\dfrac{\sin\:\theta\:.\:\cos\:\theta\:+\:\sin\:\theta}{\cos\:\theta}}{\dfrac{\sin\:\theta\:\times\:\sin\:\theta}{\cos\:\theta}}\:\right)}$

$\displaystyle{\implies\sf\:LHS\:=\:\sin\:\theta\:\left(\:\dfrac{\sin\:\theta\:.\:\cos\:\theta\:+\:\sin\:\theta}{\cancel{\cos\:\theta}}\:\times\:\dfrac{\cancel{\cos\:\theta}}{\sin\:\theta\:\times\:\sin\:\theta}\:\right)}$

$\displaystyle{\implies\sf\:LHS\:=\:\sin\:\theta\:\left(\:\dfrac{\sin\:\theta\:.\:\cos\:\theta\:+\:\sin\:\theta}{\sin\:\theta\:\times\:\sin\:\theta}\:\right)}$

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

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

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\cancel{\sin\:\theta}\:(\:\cos\:\theta\:+\:1\:)}{\cancel{\sin\:\theta}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\cos\:\theta\:+\:1}$

$\displaystyle{\sf\:RHS\:=\:\cos\:\theta\:+\:1}$

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

Hence proved!