If sin theta = m/n, then evaluate tan theta + 4 / 4cot theta + 1
Answers
Explanation:
=n2−m2m
Given:
\frac{\tan \theta+4}{4 \cot \theta+1}4cotθ+1tanθ+4
Solution:
\frac{\tan \theta + 4}{4 \cot \theta + 1}4cotθ+1tanθ+4
\Rightarrow \frac{\tan \theta\left(1 + \frac{4}{\tan \theta}\right)}{4 \cot \theta + 1}⇒4cotθ+1tanθ(1+tanθ4)
\Rightarrow \frac{\tan \theta(1 + 4 \cot \theta)}{4 \cot \theta + 1}⇒4cotθ+1tanθ(1+4cotθ)
\Rightarrow \tan \theta⇒tanθ
We know that,
\tan \theta=\frac{\sin \theta}{\cos \theta}tanθ=cosθsinθ
\Rightarrow \frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}⇒1−sin2θsinθ
From question,
\sin \theta=\frac{m}{n}sinθ=nm
\Rightarrow \frac{\frac{m}{n}}{\sqrt{1 - \left(\frac{m}{n}\right)^{2}}}⇒1−(nm)2nm
\Rightarrow \frac{\frac{m}{n}}{\sqrt{\frac{n^{2} - m^{2}}{n^{2}}}}⇒n2n2−m2nm
\Rightarrow \frac{\frac{m}{n}}{\sqrt{\frac{(n - m)(n + m)}{n^{2}}}}⇒n2(n−m)(n+m)nm
\Rightarrow \frac{m}{n} \times \frac{n}{\sqrt{(n - m)(n + m)}}⇒nm×(n−m)(n+m)n
\Rightarrow \frac{m}{\sqrt{(n - m)(n + m)}}⇒(n−m)(n+m)m
\frac{\tan \theta + 4}{4 \cot \theta + 1}=\frac{m}{\sqrt{n^{2} - m^{2}}}4cotθ+1tanθ+4=n2−m2
Answer:
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)
Explanation: