Math, asked by Deebendran, 5 months ago

If 5tan=4, find he value of 5sinø - 3cosø / 5sinø + 3cosø​

Answers

Answered by ravirajghadi
0

Step-by-step explanation:

tan=4/5, so triangle hypotenuse =√ 4²+5²=√41

Now put values in the expression

Answered by varadad25
9

Answer:

\displaystyle{\boxed{\red{\sf\:\dfrac{5\:\sin\:\theta\:-\:3\:\cos\:\theta}{5\:\sin\:\theta\:+\:3\:\cos\:\theta}\:=\:\dfrac{1}{7}}}}

Step-by-step-explanation:

We have given that \displaystyle{\sf\:5\:\tan\:\theta\:=\:4}.

We have to find the value of \displaystyle{\sf\:\dfrac{5\:\sin\:\theta\:-\:3\:\cos\:\theta}{5\:\sin\:\theta\:+\:3\:\cos\:\theta}}.

\displaystyle{\sf\:5\:\tan\:\theta\:=\:4\:\:\:-\:-\:[\:Given\:]}

\displaystyle{\implies\sf\:5\:\times\:\dfrac{\sin\:\theta}{\cos\:\theta}\:=\:4\:\:\:-\:-\:\left[\:\because\:\tan\:\theta\:=\:\dfrac{\sin\:\theta}{\cos\:\theta}\:\right]}

\displaystyle{\implies\sf\:\dfrac{5\:\sin\:\theta}{\cos\:\theta}\:\times\:\dfrac{1}{3}\:=\:\dfrac{4}{1}\:\times\:\dfrac{1}{3}\:\:\:-\:-\:-\:\left[\:Multiplying\:both\:sides\:by\:\dfrac{1}{3}\:\right]}

\displaystyle{\implies\sf\:\dfrac{5\:\sin\:\theta}{3\:\cos\:\theta}\:=\:\dfrac{4}{3}}

\displaystyle{\implies\sf\:\dfrac{5\:\sin\:\theta\:+\:3\:\cos\:\theta}{5\:\sin\:\theta\:-\:3\:\cos\:\theta}\:=\:\dfrac{4\:+\:3}{4\:-\:3}\:\:\:-\:-\:[\:By\:using\:Componendo\:-\:dividendo\:]}

\displaystyle{\implies\sf\:\dfrac{5\:\sin\:\theta\:+\:3\:\cos\:\theta}{5\:\sin\:\theta\:-\:3\:\cos\:\theta}\:=\:\dfrac{7}{1}}

\displaystyle{\implies\sf\:\dfrac{5\:\sin\:\theta\:+\:3\cos\:\theta}{5\:\sin\:\theta\:-\:3\:\cos\:\theta}\:=\:7}

\displaystyle{\implies\boxed{\red{\sf\:\dfrac{5\:\sin\:\theta\:-\:3\:\cos\:\theta}{5\:\sin\:\theta\:+\:3\:\cos\:\theta}\:=\:\dfrac{1}{7}}}\sf\:\:\:-\:-\:[\:By\:using\:Invertendo\:]}

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Additional Information:

\displaystyle{\boxed{\begin{minipage}{4.5 cm} \underline{\bf\:Trigonometric\:Identities}\\\\\sf\:1.\:\tan\:\theta\:=\:\dfrac{\sin\:\theta}{\cos\:\theta}\\\\\sf\:2.\:\sin^2\:\theta\:+\:\cos^2\:\theta\:=\:1\\\\\sf\:3.\:1\:+\:\cot^2\:\theta\:=\:\sf\:cosec^2\:\theta\\\\\sf\:4.\:1\:+\:\tan^2\:\theta\:=\:\sec^2\:\theta\end{minipage}}}

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