Login/Sign Up
Home » Prove the following identities: (i) (ii).
Prove the following identities: (i) {{\sin }^{2}}B{{\cos }^{2}}B-{{\cos }^{2}}B{{\sin }^{2}}B={{\sin }^{2}}B-{{\sin }^{2}}B(ii). {{\left( 1-\tan B \right)}^{2}}+{{\left( 1+\tan B \right)}^{2}}=2{{\sec }^{2}}B
Class 10 | Frank | ICSE | Maths | Trigonometric Identities
Prove the following identities: (i) {{\sin }^{2}}B{{\cos }^{2}}B-{{\cos }^{2}}B{{\sin }^{2}}B={{\sin }^{2}}B-{{\sin }^{2}}B(ii). {{\left( 1-\tan B \right)}^{2}}+{{\left( 1+\tan B \right)}^{2}}=2{{\sec }^{2}}B

(i)

From the question first we consider Left Hand Side (LHS),

={{\sin }^{2}}B{{\cos }^{2}}B-{{\cos }^{2}}B{{\sin }^{2}}B

We know that, {{\cos }^{2}}B=\left( 1-{{\sin }^{2}}B \right)

={{\sin }^{2}}B\left( 1-{{\sin }^{2}}B \right)-\left( 1-{{\sin }^{2}}B \right){{\sin }^{2}}B

={{\sin }^{2}}B-{{\sin }^{2}}B{{\sin }^{2}}B-{{\sin }^{2}}B+{{\sin }^{2}}B{{\sin }^{2}}B

On simplification we get,

={{\sin }^{2}}B-{{\sin }^{2}}B

Then, Right Hand Side (RHS) ={{\sin }^{2}}B-{{\sin }^{2}}B

Therefore, LHS = RHS

(ii)

From the question first we consider Left Hand Side (LHS),

={{\left( 1-\tan B \right)}^{2}}+{{\left( 1+\tan B \right)}^{2}}

Expanding the above terms we get,

=1+{{\tan }^{2}}B-2\tan A+1+{{\tan }^{2}}B+2\tan B

=2\left( 1+{{\tan }^{2}}B \right)

=2{{\sec }^{2}}B

Then, Right Hand Side (RHS) =2{{\sec }^{2}}B

Therefore, LHS = RHS

i

More Material