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Prove the following identities: (i) ${{\sin }^{2}}B{{\cos }^{2}}B-{{\cos }^{2}}B{{\sin }^{2}}B={{\sin }^{2}}B-{{\sin }^{2}}B$(ii). ${{\left( 1-\tan B \right)}^{2}}+{{\left( 1+\tan B \right)}^{2}}=2{{\sec }^{2}}B$
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

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