Question

tan 0+ sin 0/tan 0-sin 0
prove: sec0+1/sec0 -1​

Answer 1

Question:

Prove that:

$\displaystyle{\sf\:\dfrac{\tan\:\theta\:+\:\sin\:\theta}{\tan\:\theta\:-\:\sin\:\theta}\:=\:\dfrac{\sec\:\theta\:+\:1}{\sec\:\theta\:-\:1}}$

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{\tan\:\theta\:+\:\sin\:\theta}{\tan\:\theta\:-\:\sin\:\theta}\:=\:\dfrac{\sec\:\theta\:+\:1}{\sec\:\theta\:-\:1}}}}$

Step-by-step-explanation:

We have given a trigonometric equation.

We have to prove that equation.

Now,

$\displaystyle{\sf\:\dfrac{\tan\:\theta\:+\:\sin\:\theta}{\tan\:\theta\:-\:\sin\:\theta}\:=\:\dfrac{\sec\:\theta\:+\:1}{\sec\:\theta\:-\:1}}$

$\displaystyle{\sf\:RHS\:=\:\dfrac{\sec\:\theta\:+\:1}{\sec\:\theta\:-\:1}}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\dfrac{1}{\cos\:\theta}\:+\:1}{\dfrac{1}{\cos\:\theta}\:-\:1}\:\:\:-\:-\:-\:\left[\:\because\:\sec\:\theta\:=\:\dfrac{1}{\cos\:\theta}\:\right]}$

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

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{1\:+\:\cos\:\theta}{\cancel{\cos\:\theta}}\:\times\:\dfrac{\cancel{\cos\:\theta}}{1\:-\:\cos\:\theta}}$

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

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{1\:+\:\dfrac{\sin\:\theta}{\tan\:\theta}}{1\:-\:\dfrac{\sin\:\theta}{\tan\:\theta}}\:\:\:\:-\:-\:-\:\left[\:\because\:\tan\:\theta\:=\:\dfrac{\sin\:\theta}{\cos\:\theta}\:\right]}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\dfrac{\tan\:\theta\:+\:\sin\:\theta}{\tan\:\theta}}{\dfrac{\tan\:\theta\:-\:\sin\:\theta}{\tan\:\theta}}}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\tan\:\theta\:+\:\sin\:\theta}{\cancel{\tan\:\theta}}\:\times\:\dfrac{\cancel{\tan\:\theta}}{\tan\:\theta\:-\:\sin\:\theta}}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\tan\:\theta\:+\:\sin\:\theta}{\tan\:\theta\:-\:\sin\:\theta}}$

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

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

Hence proved!