Question

show that (sintheta +costheta)whole square+(sintheta-costheta)whole square=2​

Answer 1

Answer:

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Answer 2

Given:

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

To Find:

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

Solution:

Let’s take L.H.S data and find the R.H.S

Step 1:

$\begin{array}{l}{=(\sin \theta+\cos \theta)^{2}+(\sin \theta-\cos \theta)^{2}} \\ {=\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta+\sin ^{2} \theta+\cos ^{2} \theta-2 \sin \theta \cos \theta \quad \ldots \text { (i) }}\end{array}$

Step 2:

We Know $\sin ^{2} \theta+\cos ^{2} \theta=1$, Apply this in equation (i).

= 1 + 1

= 2

= R.H.S.

$\bold{\therefore(\sin \theta+\cos \theta)^{2}+(\sin \theta-\cos \theta)^{2}=2}$

L.H.S = R.H.S

Hence, proved.