Question

sin theta + cosec theta whole square + cos theta + sec theta whole square equals to 7 + 10 square theta + cot square theta

Answer 1

$\bold{(\sin \theta+\csc \theta)^{2}+(\cos \theta+\sec \theta)^{2}=7+\tan ^{2} \theta+\cot ^{2} \theta}$

Given:

$(\sin \theta+\csc \theta)^{2}+(\cos \theta+\sec \theta)^{2}=7+\tan ^{2} \theta+\cot ^{2} \theta$

To Prove:

L.H.S = R.H.S

Proof:

L.H.S

$(\sin \theta+\csc \theta)^{2}+(\cos \theta+\sec \theta)^{2}$

Using the formula of $\bold{(a+b)^{2}=\left(a^{2}+b^{2}+2 a b\right)}$ we get the value of  

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

Convert the identity of cosec to $\frac{1}{\sin \theta}$ and sec to $\frac{1}{\cos \theta}$ we get:

$=\sin ^{2} \theta+\csc ^{2} \theta+2 \sin \theta \cdot \frac{1}{\sin \theta}+\cos ^{2} \theta+\sec ^{2} \theta+2 \cos \theta \cdot \frac{1}{\cos \theta}$

Deducting common values, we get:

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

Adding the value of  $\cos ^{2} \theta+\sin ^{2} \theta=1$ in the above equation

$\begin{array}{l}{=\csc ^{2} \theta+2+1+\sec ^{2} \theta+2} \\ {=\csc ^{2} \theta+\sec ^{2} \theta+5}\end{array}$

Using the below two identities in the above step

$\csc ^{2} \theta=1+\tan ^{2} \theta$

$\sec ^{2} \theta=1+\cot ^{2} \theta$

$=\left(1+\tan ^{2} \theta+1+\cot ^{2} \theta+5\right)=\tan ^{2} \theta+\cot ^{2} \theta+7=R . H . S$

Therefore L.H.S = R.H.S

$\bold{(\sin \theta+\csc \theta)^{2}+(\cos \theta+\sec \theta)^{2}=7+\tan ^{2} \theta+\cot ^{2} \theta}$

Hence, Proved.  

Answer 2

Step-by-step explanation:

I hope this ans will help u..