Question

Prove that :

$\rm \dfrac{1}{1 + \sin ^2 θ}+ \dfrac{1}{1+ \cos ^2 θ}+ \dfrac{1}{1 + \sec ^2 θ}+ \dfrac{1}{1+ \cosec ^2 θ}=2$​

Answer 1

$\Huge\text{Refer to the explanation.}$

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

Prove that -

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

$\huge\text{\underline{\underline{Topic}}}$

$\Large\text{\{Trigonometry\}}$

$\bold{Reciprocal\ trigonometric\ ratios}$

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

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

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

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

L.H.S

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

$\bold{According\ to\ the\ reciprocal\ trigonometric\ ratios,}$

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

$\bold{By\ multiplying\ numerator\ and\ denominator,}$

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

$\bold{By\ rearrangning\ the\ denominator,}$

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

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

$=1+1$

$=2$

$=$ R.H.S

$\huge\text{\underline{\underline{Final answer}}}$

Hence, we proved that -

$\text{ \cdots\longrightarrow\boxed{\dfrac{1}{1+\sin^{2}\theta}+\dfrac{1}{1+\cos^{2}\theta}+\dfrac{1}{1+\sec^{2}\theta}+\dfrac{1}{1+\csc^{2}\theta}=2.} }$