Question

if sin theta = 3/4 , prove that whole root cosec^theta - cot^theta / sec^theta - 1 =root 7 / 3

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Answer 1

Step-by-step explanation:

We can solve this in two ways, either by written proof or visual proof,

Let's see both,

Method 1 (Written proof)

We have,

Sin θ = (3/4)

We know that,

Sin² θ + Cos² θ = 1

So,

Cos² θ = 1 - Sin² θ

Cos² θ = 1 - (3/4)²

Cos² θ = 1 - (9/16)

Cos² θ = (16/16) - (9/16)

Cos² θ = (16 - 9)/16

Cos² θ = 7/16

Cos θ = √(7/16)

Cos θ = √7/√16

Cos θ = √7/4

Now,

√((Cosec² θ - Cot² θ)/(Sec² θ - 1))

We know that,

1 + Cot² θ = Cosec² θ

Cosec² θ - Cot² θ = 1 ------ 1

Also,

1 + Tan² θ = Sec² θ

Sec² θ - 1 = Tan² θ ------ 2

Hence,

From eq.1 and eq.2 we get,

√((Cosec² θ - Cot² θ)/(Sec² θ - 1)) = √(1/Tan² θ)

We know that,

Tan θ = Sin θ/Cos θ

Tan θ = (3/4)/(√7/4)

Tan θ = 3/√7

Now,

√(1/Tan² θ) = √(1/(Tan θ)²)

= √(1/(3/√7)²

= √(1/(9/7))

= √(7/9)

= (√7)/3

Hence proved.

Method 2 (Visual Proof)

(Do refer the above image)

Now, we know that,

Sin θ = (3/4)

So,

Let AB = 3k and AC = 4k

Then,

By Pythagoras theorem,

AB² + BC² = AC²

(3k)² + BC² = (4k²)

BC² = (4k²) - (3k²)

BC² = 16k² - 9k²

BC² = 7k²

BC = √7k²

BC = k√7

Now,

√((Cosec² θ - Cot² θ)/(Sec² θ - 1))

We know that,

1 + Cot² θ = Cosec² θ

Cosec² θ - Cot² θ = 1 ------ 1

Also,

1 + Tan² θ = Sec² θ

Sec² θ - 1 = Tan² θ ------ 2

Hence,

From eq.1 and eq.2 we get,

√((Cosec² θ - Cot² θ)/(Sec² θ - 1)) = √(1/Tan² θ)

Now,

Tan θ = (3k/(k√7))

Canceling k, we get,

Tan θ = (3/√7)

Now,

Tan² θ = (3/√7)²

Tan² θ = (9/7)

So,

√(1/Tan² θ) = √(1/(9/7))

= √(7/9)

= √7/√9

= (√7)/3

Hence proved.

Hope it helped and believing you understood it........All the best