Question

If sin theta = m/n, Find the value of tan theta +4/4cot theta + 1

Answer 1

sinθ=m/n
∴, cosθ=√1-sin²θ=√1-m²/n²=√(n²-m²)/n²=√(n²-m²)/n
∴,tanθ=sinθ/cosθ=m/√(n²-m²)
cotθ=1/tanθ=√(n²-m²)/m
∴, (tanθ+4)/(4cotθ+1)
=[m/√(n²-m²)+4]/[4√(n²-m²)/m+1]
=[{m+4√(n²-m²)}/√(n²-m²)]/[{4√(n²-m²)+m}/m]
={m+4√(n²-m²)}/√(n²-m²)×m/{m+4√(n²-m²}
=m/√(n²-m²)

Answer 2

$\frac{\tan \theta+4}{4 \cot \theta+1}=\frac{m}{\sqrt{n^{2}-m^{2}}}$

Given:

$\frac{\tan \theta+4}{4 \cot \theta+1}$

Solution:

$\frac{\tan \theta + 4}{4 \cot \theta + 1}$

$\Rightarrow \frac{\tan \theta\left(1 + \frac{4}{\tan \theta}\right)}{4 \cot \theta + 1}$

$\Rightarrow \frac{\tan \theta(1 + 4 \cot \theta)}{4 \cot \theta + 1}$

$\Rightarrow \tan \theta$

We know that,

$\tan \theta=\frac{\sin \theta}{\cos \theta}$

$\Rightarrow \frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}$

From question,

$\sin \theta=\frac{m}{n}$

$\Rightarrow \frac{\frac{m}{n}}{\sqrt{1 - \left(\frac{m}{n}\right)^{2}}}$

$\Rightarrow \frac{\frac{m}{n}}{\sqrt{\frac{n^{2} - m^{2}}{n^{2}}}}$

$\Rightarrow \frac{\frac{m}{n}}{\sqrt{\frac{(n - m)(n + m)}{n^{2}}}}$

$\Rightarrow \frac{m}{n} \times \frac{n}{\sqrt{(n - m)(n + m)}}$

$\Rightarrow \frac{m}{\sqrt{(n - m)(n + m)}}$

$\frac{\tan \theta + 4}{4 \cot \theta + 1}=\frac{m}{\sqrt{n^{2} - m^{2}}}$