Login/Sign Up
Home » Prove the following identities: (iii) (iv)
Prove the following identities: (iii) \cos e{{c}^{2}}B-\cos e{{c}^{2}}B={{\cot }^{4}}B+{{\cot }^{4}}B (iv) {{\sec }^{2}}A+\cos e{{c}^{2}}A={{\sec }^{2}}A\cos e{{c}^{2}}A
Class 10 | Frank | ICSE | Maths | Trigonometric Identities
Prove the following identities: (iii) \cos e{{c}^{2}}B-\cos e{{c}^{2}}B={{\cot }^{4}}B+{{\cot }^{4}}B (iv) {{\sec }^{2}}A+\cos e{{c}^{2}}A={{\sec }^{2}}A\cos e{{c}^{2}}A

(iii)

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

=\cos e{{c}^{4}}B-\cos e{{c}^{4}}B

By taking common we get,

=\cos e{{c}^{4}}B\left( \cos e{{c}^{2}}B-1 \right)

Now we consider Right Hand Side (RHS) ={{\cot }^{4}}B+{{\cot }^{4}}B

Again taking common we get,

={{\cot }^{2}}A\left( {{\cot }^{2}}A+1 \right)

We know that, {{\cot }^{2}}B=\cos e{{c}^{2}}B-1,{{\cot }^{2}}B+1=\cos e{{c}^{2}}B

So, \left( \cos e{{c}^{2}}B-1 \right)\cos e{{c}^{2}}B

Therefore, LHS = RHS

(iv)

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

{{\sec }^{2}}A+\cos e{{c}^{2}}A

We know that, {{\sec }^{2}}A=1/{{\cos }^{2}}A,\cos e{{c}^{2}}A=1/{{\sin }^{2}}A

=\left( 1/{{\cos }^{2}}A \right)+\left( 1/{{\sin }^{2}}A \right)

=\left( {{\sin }^{2}}A+{{\cos }^{2}}A \right)/\left( {{\cos }^{2}}A{{\sin }^{2}}A \right)

=1/\left( {{\cos }^{2}}A{{\sin }^{2}} \right)

={{\sec }^{2}}A\cos e{{c}^{2}}A

Then, Right Hand Side (RHS) ={{\sec }^{2}}A\cos e{{c}^{2}}A

Therefore, LHS = RHS

i

More Material