Question

(seca+cos a)(seca- cosa)=tan^2a+sin^2a​

Answer 1

LEFT HAND SIDE :-

(secA + cosA) (secA-cosA)

We have here the identity :-

(a+b) (a-b) =a²-b²

Applying the same we have :-

= sec²A - cos²A

=1 + tan²A - 1 + sin²A

=tan²A + sin²A

=RHS

KEY POINTS TO REMEMBER :-

☸️ 1 + tan²A = sec²A

☸️ sin²A + cos²A = 1

☸️ 1 + cot²A = cosec²A

☸️ cos²A = 1 - sin²A

Since,

We proved LHS = RHS

Hence, solved!

Answer 2

Step-by-step explanation:

LHS

(seca + cos a )( Sec a -cos a)

[(a+b ) (a-b) ] = a^2- b^2

So

sec^2a -cos^2a. eq1

{sec^2a = 1+tan^2a, cos^2a = 1-sin^2a}

Putting the value in eq 1

1+tan^2 a - ( 1-sin^2 a)

1+tan^2 a- 1 +sin^2 a

tan^2 +sin^2 a

LHS = RHS

Hence proved