Question

Prove that cos^2 A - sin^2A = 2 cos^2 A - 1​

Answer 1

First, we know sin²θ+cos²θ=1.

Then, we will use equality to lead the following expressions.

→ sin²θ+cos²θ=1

→ -sin²θ-cos²θ=-1

→ (2cos²θ) - sin²θ - cos²θ = (2cos²θ) - 1

→ cos²θ - sin²θ = 2cos²θ - 1 (Proved.)

Answer 2

Given :-

$\sf cos^2A-sin^2A= 2cos^2A-1$

We have to prove the give equation-

• Identity used:-

$\boxed{\sf\ sin^2A+cos^2A=1 }$

★Solution :-

L.H.S

$:\implies\sf cos^2A-sin^2A\\ \\ \bullet\sf we \ can \ write \ sin^2A \ as [ sin^2A+cos^2=1] \\ \\ \bullet\sf \ 1- cos^2A\\ \\ :\implies\sf cos^2A-(1-cos^2A)\\ \\ :\implies\sf cos^2A-1+cos^2A\\ \\ :\implies\sf 2cos^2A-1 \\ \\ :\implies\sf 2cos^2A-1= 2cos^2A-1\\ \\ \sf \ L.H.S= R.H.S \ \ (hence \ proved \ ! )$

$\underline{\bigstar{\sf\ Other \ Trigonometric\ Identities :- }}$

$\underline{\bullet{\sf\ Pythagorean\ Identity :- }}$

$\star\sf\ \ 1+tan^2A= sec^2A\\ \\ \star\sf \ \ 1+cot^2A= cosec^2A$