Question

Show that (sin theta + cos theta)²– (sin theta - cos theta)² = 2 sin 2 theta​

Answer 1

Answer:

RHS is wrong I felt please correct it

Answer 2

Step-by-step explanation:

LHS:-

$\qquad\qquad\sf:\rightarrowtail (sin\theta+cos\theta)^2-(sin\theta-cos\theta)^2$

• using algebraic identities

$\boxed{\sf (a+b)^2=a^2+2ab+b^2}$

$\boxed{\sf (a-b)^2=a^2-2ab+b^2}$

$\qquad\qquad\sf:\rightarrowtail (sin^2\theta+cos^2\theta+2\times sin\theta\times cos\theta)-(sin^2\theta+cos^2\theta-2\times sin\theta\times cos\theta)$

$\qquad\qquad\sf:\rightarrowtail sin^2\theta+cos^2\theta+2\times sin\theta\times cos\theta-sin^2\theta-cos^2\theta+2sin\theta.cos\theta$

$\qquad\qquad\sf:\rightarrowtail sin^2\theta-sin^2\theta+cos^2\theta-cos^2\theta+2sin\theta.cos\theta+2sin\theta.cos\theta$

$\qquad\qquad\sf:\rightarrowtail 0+0+4sin\theta.cos\theta$

$\qquad\qquad\sf:\rightarrowtail 4sin\theta.cos\theta$

$\qquad\qquad\sf:\rightarrowtail RHS$

$\therefore\underline{\sf Hence\:proved.}$