If tan θ = 1/root7, then show that = (cosec sq theta

Q) If tan θ = $\frac{1}{\sqrt7}$ , then show that = $\frac {\csc^2 \theta - \sec^2 \theta}{\csc^2 \theta + \sec^2 \theta} = \frac{3}{4}$

Ans: Given that, tan θ = $\frac{1}{\sqrt7}$

$\therefore$ cot θ = √7

Let’s start from numerator of LHS:

cosec² θ – sec² θ = (1 + cot² θ) – (1 + tan² θ)

= cot² θ – tan² θ

= (√7)² – $(\frac{1}{\sqrt 7})^2$

= 7 – $\frac{1}{7}$

= $\frac{48}{7}$ ………………… (i)

Similarly, let’s solve denominator of LHS:

cosec² θ + sec² θ

= (1 + cot² θ) + (1 + tan² θ)

= 2 + cot² θ + tan² θ

= 2 + ( $\sqrt 7)^2 + (\frac{1}{\sqrt 7})^2$

= 9 + $\frac{1}{7}$

= $\frac{64}{7}$ ………………….. (ii)

Now, let’s put the values from equation (i) and equation (ii) in LHS, we get:

LHS = $\frac {\frac{48}{7}}{\frac{64}{7}}$

= $\frac {48}{64}$

= $\frac {3}{4}$ ….. RHS…. Hence Proved !

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