# (solved)Question 21 SSC-CGL 2020 March 3 Shift 1

If (secθ + tanθ)/(secθ - tanθ) = 5/3, what is the value of (cosecθ + cotθ)/(cosecθ - cotθ)?
(Rev. 20-Jan-2024)

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### Question 21SSC-CGL 2020 Mar 3 Shift 1

If (secθ + tanθ)/(secθ - tanθ) = 5/3, what is the value of (cosecθ + cotθ)/(cosecθ - cotθ)?

### Solution in Brief

Divide numerator and denominator of (secθ + tanθ)/(secθ - tanθ) by secθ to get cosecθ = 4, and then use cosec² θ- cot² θ = 1, to get cotθ = √15. Put these in the required expression to get 31 + 8√15 answer!

### Solution in Detail

Given $\displaystyle \frac{\sec \theta + \tan \theta }{\sec \theta - \tan \theta } = \frac 53$

Divide num and den of LHS by $\displaystyle \tan \theta$

$\displaystyle \therefore \frac{\cosec \theta + 1}{\cosec \theta - 1} = \frac 53$

REMEMBER: componendo dividendo?

if $\displaystyle \frac{p + q}{p - q} = \frac ab$, then $\displaystyle \frac pq = \frac{a + b}{a - b}$

$\displaystyle \therefore \cosec \theta = \frac{5 + 3}{5 - 3} = 4$

Take $\displaystyle \theta$ as an acute angle

But $\displaystyle \cot^2 \theta = \cosec^2 \theta - 1$

$\displaystyle \therefore \cot^2 \theta = 4^2 - 1$

$\displaystyle \therefore \cot \theta = \sqrt {15}$

$\displaystyle \frac{\cosec \theta + \cot \theta}{\cosec \theta - \cot \theta} = \frac{4 + \sqrt {15}}{4 - \sqrt {15}}$

Rationalizing, we get

$\displaystyle = \frac{4 + \sqrt {15}}{4 - \sqrt {15}} \times \frac{4 + \sqrt {15}}{4 + \sqrt {15}}$

$\displaystyle = 31 + 8\sqrt{15} \:\underline{Ans}$