Question

cosec2Q+sec2Q=cosec2Q sec2Q

Answer 1

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