Math, asked by ishikapatidar6216, 1 year ago

Sinq +cosQ/(sinQ-cosQ)+sinQ-cosQ/sinq+cosq=2/2sin2Q-1

Answers

Answered by aquialaska
9

Answer:

To prove: \frac{sin\,Q\,+\,cos\,Q}{sin\,Q\,-\,cos\,Q}+\frac{sin\,Q\,-\,cos\,Q}{sin\,Q\,+\,cos\,Q}=\frac{2}{2sin^2\,Q\,-\,1}

Proof,

Consider

LHS  =\,\frac{sin\,Q\,+\,cos\,Q}{sin\,Q\,-\,cos\,Q}+\frac{sin\,Q\,-\,cos\,Q}{sin\,Q\,+\,cos\,Q}

        =\,\frac{(sin\,Q\,+\,cos\,Q)^2\,+\,(sin\,Q\,-\,cos\,Q)^2}{(sin\,Q\,-\,cos\,Q)(sin\,Q\,+\,cos\,Q)}

Using identity, (x+y)^2=x^2+y^2+2xy\:,\:(x-y)^2=x^2+y^2-2xy\:and\:(x-y)(x+y)=x^2-y^2  we get,

        =\,\frac{(sin\,Q)^2\,+\,(cos\,Q)^2\,+\,2\,sin\,Q\,cos\,Q\,+\,(sin\,Q)^2\,+\,(cos\,Q)^2\,-\,2\,sin\,Q\,cos\,Q}{(sin\,Q)^2\,-\,(cos\,Q)^2}

        =\,\frac{(sin\,Q)^2\,+\,(cos\,Q)^2\,+\,(sin\,Q)^2\,+\,(cos\,Q)^2}{-\,((cos\,Q)^2\,-\,(sin\,Q)^2\,)}

Now using trigonometric identity sin^2\theta+cos^2\theta=1\:and\:cos\,2\theta=1-2\,sin^2\theta  we get

        =\,\frac{1\,+\,1}{-\,(1-2\,sin^2\,Q)}

        =\,\frac{2}{2\,sin^2\,Q-1}

      = RHS

Hence Proved

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