Question

proove that 1 upon sec theta + tan theta = 1- sin theta upon cos theta​

Answer 1

Step-by-step explanation:

LHS

• 1/sec¢ + tan¢ [¢=theta]
• 1/sec¢ + tan¢ × sec¢ - tan¢/sec¢ - tan¢ [rationalising it]
• sec¢ - tan¢ / sec²¢- tan²¢
• sec¢ - tan¢/ 1 [sec²¢-tan²¢=1]
• 1/cos¢ - sin¢/ cos¢ [sec¢=1/cos¢ & tan¢=sin¢/cos¢]

• 1-sin¢ / cos¢

LHS=RHS

Hence, proved^__^

hope it helps...

Answer 2

Question:

Prove that

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

Answer:

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

Step-by-step-explanation:

We have a trigonometric equation.

We have to prove that equation.

The given trigonometric equation is

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

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

We know that,

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

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

$\displaystyle{\implies\sf\:RHS\:=\:(\:1\:-\:\sin\:\theta\:)\:\times\:\sec\:\theta}$

$\displaystyle{\implies\sf\:RHS\:=\:\sec\:\theta\:-\:\sin\:\theta\:\times\:\sec\:\theta}$

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

$\displaystyle{\implies\sf\:RHS\:=\:\sec\:\theta\:-\:\dfrac{\sin\:\theta}{\cos\:\theta}}$

We know that,

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

$\displaystyle{\implies\sf\:RHS\:=\:\sec\:\theta\:-\:\tan\:\theta}$

Multiplying and dividing by $\displaystyle{\sf\:\sec\:\theta\:+\:\tan\:\theta}$ we get,

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

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{(\:\sec\:\theta\:-\:\tan\:\theta\:)\:(\:\sec\:\theta\:+\:\tan\:\theta\:)}{\sec\:\theta\:+\:\tan\:\theta}}$

We know that,

$\displaystyle{\boxed{\green{\sf\:(\:a\:+\:b\:)\:(\:a\:-\:b\:)\:=\:a^2\:-\:b^2\:}}}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\sec^2\:\theta\:-\:\tan^2\:\theta}{\sec\:\theta\:+\:\tan\:\theta}}$

We know that,

$\displaystyle{\boxed{\orange{\sf\:\sec^2\:\theta\:=\:1\:+\:\tan^2\:\theta\:}}}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{1\:+\:\cancel{\tan^2\:\theta}\:-\:\cancel{\tan^2\:\theta}}{\sec\:\theta\:+\:\tan\:\theta}}$

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

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

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

Hence proved!