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Prove the following identities: (v) {{\cos }^{4}}A-{{\sin }^{4}}A=2{{\cos }^{2}}A-1 (vi) \left( \sec B-\cos B \right)\left( \sec B+\cos B \right)={{\sin }^{2}}B+{{\tan }^{2}}B
Class 10 | Frank | ICSE | Maths | Trigonometric Identities
Prove the following identities: (v) {{\cos }^{4}}A-{{\sin }^{4}}A=2{{\cos }^{2}}A-1 (vi) \left( \sec B-\cos B \right)\left( \sec B+\cos B \right)={{\sin }^{2}}B+{{\tan }^{2}}B

(v)

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

={{\cos }^{4}}A-{{\sin }^{4}}A

We know that, {{a}^{2}}-{{A}^{2}}=\left( a+A \right)\left( a-A \right)

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

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

=2{{\cos }^{2}}A-1

Then, Right Hand Side (RHS) =2{{\cos }^{2}}A-1

Hence, LHS = RHS

(vi)

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

=(secB-cosB)(secB+cosB)

We know that, {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)

=\left( {{\sec }^{2}}B-{{\cos }^{2}}B \right)

Blso we know that, {{\sec }^{2}}B=1+{{\tan }^{2}}B,{{\cos }^{2}}B=1-{{\sin }^{2}}B

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

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

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

Therefore, LHS = RHS

i

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