Math, asked by rp12, 1 year ago

cosec2Q+sec2Q=cosec2Q sec2Q

Answers

Answered by harendrachoubay
2

\csc^2 \theta +\sec^2 \theta =\csc^2 \theta.\sec^2 \theta, proved.

Step-by-step explanation:

To prove that, \csc^2 \theta +\sec^2 \theta =\csc^2 \theta.\sec^2 \theta

L.H.S. =\csc^2 \theta +\sec^2 \theta

Using the trigonometric identity,

\csc^2 \theta =1+\cot^2 \theta and

\sec^2 \theta =1+\tan^2 \theta

= 1+\cot^2 \theta+1+\tan^2 \theta

=2+\cot^2 \theta +\tan^2 \theta

= 2+\dfrac{\cos^2 \theta}{\sin^2 \theta} +\dfrac{\sin^2 \theta}{\cos^2 \theta}

Taking LCM sin^2 \theta\cos^2 \theta, we get

=\dfrac{2\sin^2 \theta\cos^2 \theta+(\cos^2 \theta)^{2}+(\sin^2 \theta)^{2}  }{\sin^2 \theta\cos^2 \theta}

=\dfrac{(\sin^2 \theta)^{2}+(\cos^2 \theta)^{2}+2\sin^2 \theta\cos^2 \theta}{\sin^2 \theta\cos^2 \theta}

Using the algebraic identity,

(a+b)^{2}=a^{2}+b^{2} +2ab

= \dfrac{(\sin^2 \theta+\cos^2 \theta)^{2}}{\sin^2 \theta\cos^2 \theta}

Using the trigonometric identity,

\sin^2 \theta +\cos^2 \theta=1

= \dfrac{(1)^{2}}{\sin^2 \theta\cos^2 \theta}

Using the trigonometric identity,

\csc \theta= \dfrac{1}{\sin \theta} and

\sec \theta= \dfrac{1}{\cos \theta}

=\csc^2 \theta.\sec^2 \theta

= R.H.S., proved.

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