If $\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}=\frac{3}{5}$, then the value of $\frac{{cosec} \theta+\cot \theta}{{cosec} \theta-\cot \theta}$ is :
Answer & explanation
Correct answer: option 2
\(\frac{secθ - tanθ}{secθ + tanθ}\) = \(\frac{3}{5}\)
By using componendo and dividendo
\(\frac{secθ }{ tanθ}\) = \(\frac{3 +5}{5 - 3}\)
\(\frac{secθ }{ tanθ}\) = \(\frac{8}{2}\)
\(\frac{1 }{sinθ}\) = \(\frac{4}{1}\)
sinθ = \(\frac{1}{4}\)
{ sinθ = \(\frac{P}{H}\) }
By using pythagoras theorem,
P² + B² = H²
1² + B² = 4²
B = \(\sqrt {15 }\)
Now,
\(\frac{cosecθ+ cotθ}{cosecθ- cotθ}\)
= \(\frac{H+ B}{H- B}\)
= \(\frac{4+ √15}{4- √15}\)
= \(\frac{4+ √15}{4- √15}\) x \(\frac{4+ √15}{4+ √15}\)
= (4+ √15)²
= 16 + 15 + 8√15
= 31 + 8√15