Math, asked by rameshsarupuru, 3 months ago

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

Answers

Answered by charapalesakshi
0

Answer:

RHS is wrong I felt please correct it

Answered by NewGeneEinstein
6

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.}

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