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27.$\frac{{{(1+\sin \theta )}^{2}}+{{(1-\sin \theta )}^{2}}}{2{{\cos }^{2}}\theta }=\frac{1+{{\sin }^{2}}\theta }{1-{{\sin }^{2}}\theta }$
27.$\frac{{{(1+\sin \theta )}^{2}}+{{(1-\sin \theta )}^{2}}}{2{{\cos }^{2}}\theta }=\frac{1+{{\sin }^{2}}\theta }{1-{{\sin }^{2}}\theta }$
CBSE | Class 10 | Exercise 6.1 | Maths | RD Sharma | Trigonometric Identities
27.$\frac{{{(1+\sin \theta )}^{2}}+{{(1-\sin \theta )}^{2}}}{2{{\cos }^{2}}\theta }=\frac{1+{{\sin }^{2}}\theta }{1-{{\sin }^{2}}\theta }$

Firstly we will solve L.H.S=R.H.S

Then use trigonometric identity $\sin \theta +\cos \theta =1,$, we get

$LHS=\frac{{{(1+\sin \theta )}^{2}}+{{(1-\sin \theta )}^{2}}}{2{{\sec }^{2}}\theta }$

$=\frac{(1+2\sin \theta +{{\sin }^{2}}\theta )+(1-2\sin \theta +{{\sin }^{2}}\theta )}{2{{\cos }^{2}}\theta }$

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

$=\frac{2+2{{\sin }^{2}}\theta }{2{{\cos }^{2}}\theta }$

$=\frac{2(1+{{\sin }^{2}}\theta )}{2{{\cos }^{2}}\theta }$

$=\frac{(1+{{\sin }^{2}}\theta )}{(1-{{\sin }^{2}}\theta )}$

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

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