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RD Chapter 5 Trigonometric Functions Ex 5.1 Solutions

Question - 11 : -

Answer - 11 : -

Let us consider LHS:

By using the formula,

tan θ = sin θ / cos θ;

cot θ = cos θ / sin θ

Now,

By using the formula, a3 + b3 =(a + b) (a2 + b2– ab)

We know, sinx + cosx= 1.

1 – 1 + sin x cos x

Sin x cos x

= RHS

LHS = RHS

Hence proved.

Question - 12 : -

Answer - 12 : -

Let us consider LHS:

By using the formula,

cosec θ = 1/sin θ,

sec θ = 1/cos θ;

= RHS

LHS = RHS

Hence proved.

Question - 13 : - (1 + tan α tan β) 2 + (tan α – tan β) 2 = sec2 α sec2 β

Answer - 13 : -

Let us consider LHS: (1 + tan α tan β) 2 +(tan α – tan β) 2

1+ tan2 α tan2 β + 2tan α tan β + tan2 α + tan2 β – 2 tan α tan β

1 + tan2 α tan2 β + tan2 α+ tan2 β

tan2 α (tan2 β + 1) +1 (1 + tan2 β)

(1 + tan2 β) (1 + tan2 α)

We know, 1 + tan2 θ = sec2 θ

So,

sec2 α sec2 β

= RHS

LHS = RHS

Hence proved.

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