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

(i)

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

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

We know that, secA=1/cosB, tanB=sinB/cosB

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

By taking LCM we get,

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

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

Also we know that, {{\cos }^{2}}B=1-{{\sin }^{2}}B

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

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

=(1-sinB)/(1+sinB)

Then, Right Hand Side (1-sinB)/(1+sinB)

Therefore, LHS = RHS

(ii)

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

=(1/(sinB+cosB))+(1/(sinB-cosB))

By taking LCM we get,

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

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

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

Then, Right Hand Side =2\sin B/\left( 1-2{{\cos }^{2}}B \right)

Therefore, LHS = RHS

i

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