Question

sec^2 theta - cos^2 theta = sin^2 theta (sec^2 theta +1)​

Answer 1

$\large { \fcolorbox{gray}{black}{ ✔\: \textbf{Verified \: answer}}}$

$Proved \: that \sec^{2}\theta - \cos^{2}\theta = \sin^{2}\theta(\sec^{2}\theta + 1 ) \\ \\ Step-by-step \: explanation: \\ \\ We \: have \: to \: prove \: that \\  \sec^{2}\theta - \cos^{2}\theta = \sin^{2}\theta(\sec^{2}\theta + 1 ) \\ \\ Now, \: left \: hand \: side \\ \\ = \sec^{2}\theta - \cos^{2}\theta \\ = 1 + \tan^{2}\theta + \sin^{2}\theta - 1 \\ \tiny \: {Since, \: we \: know \: that  \: \sec^{2}\theta - \tan^{2}\theta = 1sec2θ−tan2θ=1 and \sin^{2}\theta + \cos^{2}\theta = 1 }\\ = \tan^{2}\theta + \sin^{2}\theta\\ = \frac{\sin^{2}\theta }{\cos^{2}\theta} + \sin^{2}\theta \\ = \sin^{2}\theta[\frac{1}{\cos^{2}\theta} + 1 ] \\ = \sin^{2}\theta(\sec^{2}\theta + 1 )\\ = right \: hand \: s ide. \\ \\ Hence, \: it \: proved.$

Answer 2

$\huge\text{\underline{\underline{Question}}}$

Prove that -

$\cdots\longrightarrow \sec^{2}\theta-\cos^{2}\theta=\sin^{2}\theta(\sec^{2}\theta+1).$

$\huge\text{\underline{\underline{Review}}}$

$\bold{Trigonometric\ identities}$

$\bullet\ \sin^{2}\theta+\cos^{2}\theta=1$

$\bullet\ \tan^{2}\theta+1=\sec^{2}\theta$

$\bold{Reciprocal\ trigonometric\ ratios}$

$\bullet\ \csc\theta=\dfrac{1}{\sin\theta}$

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

$\bullet\ \cot\theta=\dfrac{1}{\tan\theta}$

$\huge\text{\underline{\underline{Explanation}}}$

L.H.S

$=\sec^{2}\theta-\cos^{2}\theta$

$\text{ =\dfrac{1}{\cos^{2}\theta}-\cos^{2}\theta }$

$\text{ =\dfrac{1-\cos^{4}\theta}{\cos^{2}\theta} }$

$\text{ =\dfrac{(1-\cos^{2}\theta)(1+\cos\theta^{2})}{\cos^{2}\theta} }$

$\text{ =\dfrac{\sin^{2}\theta(1+\cos\theta^{2})}{\cos^{2}\theta}\ \ \ \ \ \boxed{\because\sin^{2}\theta+\cos^{2}\theta=1} }$

$\text{ =\sin^{2}\theta\times\dfrac{1+\cos\theta^{2}}{\cos^{2}\theta} }$

$\text{ =\sin^{2}\theta\times\left(1+\dfrac{1}{\cos^{2}\theta}\right) }$

$\text{ =\sin^{2}\theta\times\left(1+\sec^{2}\theta\right)\ \ \ \ \ \boxed{\because\sec\theta=\dfrac{1}{\cos\theta}} }$

$=$ R.H.S

Hence it is proved that -

$\text{ \cdots\longrightarrow\boxed{\sec^{2}\theta-\cos^{2}\theta=\sin^{2}\theta(\sec^{2}\theta+1).} }$

$\huge\text{\underline{\underline{Helpful guide}}}$

This question is related to trigonometric identities. It is better to write the equation in terms of $\sin\theta$ and $\cos\theta$.

Let us find an example.

$\cdots\longrightarrow \tan^{2}\theta+\sec^{2}\theta=2\tan^{2}\theta+1$

L.H.S

$=\tan^{2}\theta+\sec^{2}\theta$

$\text{ =\dfrac{\sin^{2}\theta}{\cos^{2}\theta}+\dfrac{1}{\cos^{2}\theta} }$

$\text{ =\dfrac{\sin^{2}\theta}{\cos^{2}\theta}+\dfrac{1}{\cos^{2}\theta} }$

$\text{ =\dfrac{\sin^{2}\theta}{\cos^{2}\theta}+\dfrac{\sin^{2}\theta+\cos^{2}\theta}{\cos^{2}\theta}\ \ \ \ \ \boxed{\because\sin^{2}\theta+\cos^{2}\theta=1} }$

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

$\text{ =\dfrac{2\sin^{2}\theta}{\cos^{2}\theta}+1 }$

$\text{ =2\tan^{2}\theta+1 \ \ \ \ \ \boxed{\because\tan\theta=\dfrac{\sin\theta}{\cos\theta}}}$

$=$ R.H.S

So, we write in terms of $\sin\theta$ and $\cos\theta$ to solve this problem.