Question

6. (1 - tan A)2 + (1 + tan A)2 = 2 sec-A​

Answer 1

Correct Question :

Prove :

$\sf{{(1 - tan \: A)}^{2} + {(1 + tan \: A)}^{2} = 2 \: {sec}^{2} \: A}$

Answer :-

Given to prove :

$\sf{{(1 - tan \: A)}^{2} + {(1 + tan \: A)}^{2} = 2 \: {sec}^{2} \: A}$

Solution :

L.H.S = $\sf{{(1 - tan \: A)}^{2} + {(1 + tan \: A)}^{2}}$

$\sf{\longrightarrow \: {(1 - tan \: A)}^{2} + {(1 + tan \: A)}^{2}}$

$\sf{\longrightarrow \: (1 + {tan \: A}^{2} \cancel{- 2 \: tanA}) \: + (1 + {tan \: A}^{2} + \cancel{2 \: tanA})}$

$\sf{\longrightarrow \: (1 + {tan \: A}^{2}) + (1 + {tan \: A}^{2})}$

$\sf{\longrightarrow \: 2(1 + {tan \: A}^{2})}$

$\sf{\longrightarrow \: 2 \: {sec}^{2} \: A} \: = R.H.S \: \: \: \: \: \: \: \: \: \: \: {}_{.. \: .. \: ..} \: \lgroup \: \sf{1 + {tan}^{2} \: A = {sec}^{2} \: A} \rgroup$

Hence, Proved!

⠀⠀⠀⠀⠀ ⠀_______________________

Remember :-

• $\sf{1 + {tan}^{2} \: A = {sec}^{2} \: A}$