Question

Please answer this fast​

Answer 1

Step-by-step explanation:

answered by ashutosh Tiwari

Answer 2

Question : -

In a right angled ∆ ABC, angle B = 90° and 15 cos A - 8 sin A = 0 . Then the value of √[({1 - cos C}{1 + cos C})/({1 - sin C}{1 + sin C})] =

ANSWER

Given : -

In a right angled ∆ ABC, angle B = 90° and 15 cos A - 8 sin A = 0 .

Required to find : -

• √[({1 - cos C}{1 + cos C})/({1 - sin C}{1 + sin C})] =

Solution : -

The equation whose value we need to evaluate is

√[({1 - cos C}{1 + cos C})/({1 - sin C}{1 + sin C})] =

• (a+b)(a-b) = a²-b²

We have;

√[({1}² - {cos C}²)/({1}² - {sin C}²)]

√[(1 - cos² C)/(1 - sin² C)]

since,

• sin² x + cos² x = 1

This implies;

• sin² x = 1 - cos² x
• cos² x = 1 - sin² x

We have;

√[(sin² C)/(cos² C)]

√[(sin C)/(cos C)]²

square and square root gets cancelled

= tan C

We have;

√[({1 - cos C}{1 + cos C})/({1 - sin C}{1 + sin C})] = tan C ....(1)

Consider this as equation 1

It is given that;

15 cos A - 8 sin A = 0

15 cos A = 8 sin A

15 = (8 sin A)/(cos A)

(15)/(8) = (sin A)/(cos A)

=> tan A = (15)/(8)

As,

• (sin x)/(cos x) = tan A

Now, By drawing a right angle triangle with angle B = 90°

we can say that;

Tan A = (opposite side)/(Adjacent side) = (BC)/(AB)

But,

Tan A = (15)/(8)

so,

BC = 15k & AB = 8k where k is a +ve integer..

Now,

Let's evaluate the value of Tan C

Tan C = (opp. side)/(adj. side)

Tan C = (AB)/(BC)

Tan C = (8k)(15k)

Tan C = (8)/(15)

Therefore,

√[({1 - cos C}{1 + cos C})/({1 - sin C}{1 + sin C})] = (8)/(15) {From eq - 1}