Question

Prove that (secA-cosA)(cotA+tanA)=tanA secA​

Answer 1

Given :

(secA-cosA)(cotA+tanA)=tanA secA

To prove :

LHS = RHS

Proof :

$\tt R.H.S = tanA \: secA$

$\tt LHS = (secA-cosA)(cotA+tanA)$

Now , we know that :-

$\boxed{ \boxed{ \tt sec \: x = \frac{1}{cos \: x} }} \\ \\ \boxed{ \boxed{ \tt cot \: x = \frac{cos \: x}{sin \: x} }} \\ \\ \boxed{ \boxed{ \tt tan \: x = \frac{sin \: x}{cos \: x} }}$

$\tt \longrightarrow( \dfrac{1}{cos \: A} -cos \: A)(\dfrac{cos \: A}{sin \: A}+\dfrac{sin \: A}{cos \: A}) \\ \\ \tt \longrightarrow( \dfrac{1- {cos}^{2} \: A}{cos \: A} )(\dfrac{ {cos}^{2} \: A + {sin}^{2} } {cos \: A \: sin \: A}) \\ \\ \tt\longrightarrow( \dfrac{ {sin}^{2} \: A}{cos \: A} )(\dfrac{1 } {cos \: A \: sin \: A}) \\ \\ \\ \tt\longrightarrow( \dfrac{ {sin} \: A}{cos \: A} )(\dfrac{1 } {cos \: A \: }) \\ \\ \\ \tt\longrightarrow \: tan \: A \: \: {sec \: A \: }$

Therefore, LHS = RHS

Hence Proved