If \(\sin \theta = \frac{8}{{17}}\) where \(\theta \) is an acute angle, then what is the value of \(\tan \theta + \cot \theta ?\)

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 09 Dec 2022 Shift 3)

Answer 1

Given:

\(\sin θ = \frac{8}{{17}}\), where θ is an acute angle

Formula Used:

sin²θ + cos²θ = 1

tan θ = \(\frac{\sinθ}{\cosθ}\) , and

cot θ = \(\frac{\cosθ}{\sinθ}\)

Calculation:

In order to find \(\tanθ\) and \(\cotθ\), we need \(\cosθ\):

sin2θ + cos2θ = 1

cos θ = \(\sqrt{1-{(\sinθ)}^2} \)

⇒ \(\sqrt{1-{(\frac{8}{17})}^2}\)

⇒ \(\sqrt{1-{(\frac{64}{289})}}\)

⇒ \(\sqrt{\frac{225}{289}}\)

⇒ \(\frac{15}{17}\)

Now, calculating \(\tan θ + \cot θ\):

⇒\(\frac{\sinθ}{\cosθ}+\frac{\cosθ}{\sinθ}\)

⇒\(\frac{(\sinθ)^2+(\cosθ)^2}{\sinθ\cosθ}\)

⇒\(\frac{1}{\sinθ\cosθ}\)

⇒\(\frac{1}{\frac{8}{17}\times\frac{15}{17}}\)

⇒\(\frac{17\times17}{8\times15}\)

⇒\(\frac{289}{120}\)

∴ The value of tan θ + cot θ is \(\bf \frac{289}{120}\).

Shortcut Trick

\(\sin θ = \frac{8}{{17}}\) = P/H, we know the Pythagorean triplet (8, 15, 17)

Then, base (B) = 15, so tan θ = P/B = 8/15, cot θ = 15/17

⇒ tan θ + cot θ

⇒ 8/15 + 15/8

⇒ \(\bf \frac{289}{120}\)