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Prove that (1+secθ-tanθ)/(1+secθ+tanθ) = (1-sinθ)/cosθ

Solution:

We have to prove that (1 + secθ - tanθ)/(1 + secθ + tanθ) = (1 - sinθ)/cosθ

Considering LHS,

LHS : (1 + secθ - tanθ)/(1 + secθ + tanθ)

By using trigonometric identity,

sec² A - tan² A = 1

(1 + secθ - tanθ) = secθ - tanθ + (sec²θ - tan²θ)

= (secθ - tanθ) + [(secθ - tanθ)(secθ + tanθ)]

Taking out common term,

= (secθ - tanθ)[1 + secθ + tanθ]

Now, (1 + secθ - tanθ)/(1 + secθ + tanθ) = (secθ - tanθ)[1 + secθ + tanθ] / (1 + secθ + tanθ)

= (secθ - tanθ)

We know that sec A = 1/cos A and tan A = sin A/cos A

So, (secθ - tanθ) = (1/cosθ - sinθ/cosθ)

= (1 - sinθ)/cosθ

= RHS

LHS = RHS

Therefore, (1 + secθ - tanθ)/(1 + secθ + tanθ) = (1 - sinθ)/cosθ

✦ Try This: Prove that : (cosecθ - sinθ)(secθ - cosθ) = 1/(tanθ + cotθ)

Considering LHS,

LHS = (cosecθ - sinθ)(secθ- cosθ)

We know that cosec A = 1/sin A and sec A = 1/cos A

So, cosecθ - sinθ = 1/sinθ - sinθ

= (1 - sin²θ)sinθ

By using trigonometric identities,

1 - sin² A = cos² A

1 - cos²A = sin²A

= cos²θ/sinθ

sec θ - cosθ = 1/cosθ - cosθ

= (1 - cos²θ)cosθ

= sin²θ/cosθ

Now, (cosecθ - sinθ)(secθ - cosθ) = (cos²θ/sinθ)(sin²θ/cosθ)

= sinθ cosθ

Considering RHS,

RHS = 1/(tanθ + cotθ)

We know that tan A = sin A/cos A and cot A = cos A/sin A

= 1/(sinθ/cosθ + cosθ/sinθ)

= 1/(sin²θ + cos²θ)/sinθcosθ

By using trigonometric identity,

cos² A + sin² A = 1

= sinθcosθ

LHS = RHS

Therefore, (cosecθ - sinθ)(secθ - cosθ) = 1/(tanθ + cotθ)

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 8

NCERT Exemplar Class 10 Maths Exercise 8.4 Problem 12

Prove that (1+secθ-tanθ)/(1+secθ+tanθ) = (1-sinθ)/cosθ

Summary:

It is proven that (1+secθ-tanθ)/(1+secθ+tanθ) = (1-sinθ)/cosθ

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