Question

5 cot theta =4, then find the value of (2 sin theta-5 cos theta)+sin theta is

Answer 1

Answer:-5/root41

Step-by-step explanation:

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Answer 2

$(2\sin\theta-5\cos\theta)+\sin\theta=-\frac{5}{\sqrt{41}}$

Step-by-step explanation:

Given:

$5 \cot\theta=4$

$\cot\theta=\frac{4}{5}$

Formula:$1+\cot^2\theta=\cosec^2\theta$

Using the formula

$1+(\frac{4}{5})^2=\cosec^2\theta$

$1+\frac{16}{25}=\cosec^2\theta$

$\frac{25+16}{25}=\cosec^2\theta$

$\cosec\theta=\sqrt{\frac{41}{25}}=\frac{\sqrt{41}}{5}$

$\sin\theta=\frac{1}{cosec\theta}=\frac{5}{\sqrt{41}}$

$\cos\theta=\sqrt{1-\sin^2\theta}$

$\cos\theta=\sqrt{1-(\frac{5}{\sqrt{41}})^2}=\sqrt{1-\frac{25}{41}}=\sqrt{\frac{41-25}{41}}=\sqrt{16}{41}}=\frac{4}{\sqrt{41}}$

$(2\sin\theta-5\cos\theta)+\sin\theta$

Substitute the values

$(2\times \frac{5}{\sqrt{41}}-5\times \frac{4}{\sqrt{41}})+\frac{5}{\sqrt{41}}$

$(\frac{10}{\sqrt{41}}-\frac{20}{\sqrt{41}})+\frac{5}{\sqrt{41}}$

$\frac{10-20}{\sqrt{41}}+\frac{5}{\sqrt{41}}\\\frac{-10}{\sqrt{41}}+\frac{5}{\sqrt{41}}$

$\frac{-10+5}{\sqrt{41}}$

$-\frac{5}{\sqrt{41}}$

#Learn more:

https://brainly.in/question/3669500:Answered by AryanTannyson