Question

you have to prove
(sin A + cos A)² + (sin A - cos A)² = 2​

Answer 1

Answer:

L.H.S

(sin A + cos A)² + (sin A - cos A)²

= sin²A + cos²B + 2 sinA cosB

+ sin²A + cos²B - 2sinA cosB

= sin²A + cos²B + sin²A + cos²B

= 1 + 1

=2

=R.H.S (proved)

Note:

* sin²A+cos²A=1

*sec²A-tan²A=1

*cosec²A-cot²A=1

(i) (a+b)²= a²+b²+2ab

(ii) (a-b)²= a²+b²-2ab

Answer 2

Given :

(sin A + cos A)² + (sin A - cos A)² = 2

To prove :

LHS = RHS

Proof :

RHS = 2

LHS = (sin A + cos A)² + (sin A - cos A)²

= (sin A + cos A)² + (sin A - cos A)²

= sin² A + cos² A +2 sin A cos A + sin² A + cos² A - 2 sin A cos A

We know that :-

sin² x + cos² y = 1

= 1 +2 sin A cos A + 1 - 2 sin A cos A

= 1+1 +2 sin A cos A - 2 sin A cos A

= 2 +0

= 2

Therefore, LHS = RHS

hence proved

Additional Information :

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$