Question

11. If tan 0 = 1√7, show that
(cosec^2 0 - sec^2 0) /(cosec^2 0+sec^2 0 3/4.​

Answer 1

Step-by-step explanation:

Given:-

tan 0 = 1/√7

To show :-

(cosec^2 0 - sec^2 0) /(cosec^2 0+sec^2 0) = 3/4

Solution:-

Given that :-

Tan 0 = 1/√7

=>Cot 0 = √7

=>Cot ^2 0 = (√7)^2

=>Cot^2 0 = 7

We know that

Cosec^2 A - Cot^2 A = 1

=>Cosec^2 A = 1 + Cot^2 A

=>Cosec^2 0 = 1+7

Cosec^2 0 = 8 -----------(1)

And we have Tan 0 = 1/√7

=>Tan^2 0 = (1/√7)^2

=>Tan^2 0 = 1/7

We know that

Sec^2 A - Tan^2 A = 1

=>Sec^2 A = 1 + Tan^2 A

=>Sec^2 0 = 1+(1/7)

=>Sec^2 0 = (7+1)/7

=>Sec^2 0 = 8/7 -----------(2)

(cosec^2 0 - sec^2 0) /(cosec^2 0+sec^2 0)

From (1) &(2)

=>[8-(8/7)]/[8+(8/7)]

=>[(56-8)/7]/[(56+8)/7]

=>(48/7)/(64/7)

=>48/64

=>(3×16)/(4×16)

=>3/4

Hence, Proved

Answer:-

(cosec^2 0 - sec^2 0) /(cosec^2 0+sec^2 0) = 3/4

Used formulae:-

• Tan A = 1/Cot A
• Cot A = 1/Tan A
• Cosec^2 A - Cot^2 A = 1
• Sec^2 A - Tan^2 A = 1