Question

If `tan θ = 1/sqrt(7)`, show that `("cosec"^2θ - sec^2θ)/("cosec"^2θ + sec^2θ) = 3/4` .

Answer 1

Let us consider a right ΔABC, right-angled at B and ∠C = θ.

Now it is given that `tan θ = (AB)/(BC) = 1/(sqrt(7))`

So, if AB = k, then BC = `sqrt(7)k`, where k is a positive number.

Using Pythagoras theorem, we have:

AC² = AB² + BC²

⇒ `AC^2 = (k)^2 + (sqrt(7K))`

⇒ AC² = k² + 7k²

⇒ `AC = 2sqrt(2k)`

Now, finding out the values of the other trigonometric ratios, we have:

`sin θ = (AB)/(AC)` 

= `k/(2sqrt(2k))`

= `1/(2sqrt(2))`

`cos θ = ( BC)/(AC)` 

= `(sqrt(7k))/(2 sqrt(2k))`

= `(sqrt(7))/(2sqrt(2))`

∴ cosec θ = `1/(sin θ) = 2 sqrt(2)` and sec θ = `1/cos θ = (2sqrt(2))/(sqrt(7))`

Substituting the values of cosec θ and sec θ in the given expression, we get:

`("cosec"^2θ - sec^2θ)/("cosec"^2θ + sec^2θ)`

=`((2 sqrt(2))^2 - ((2sqrt(2))/sqrt(7))^2)/(2sqrt((2)^2) + ((2sqrt(2))/sqrt(7))^2)`

=`(8 - (8/7))/(8 + (8/7))`

= `((56  -  8)/7)/((56  +  8)/7)`

= `48/64` = `3/4` = RHS

i.e., LHS = RHS

Hence proved.