Question

3.If 17 sin 0 = 8, then find the value of tan teta by 1+tan² teta​

Answer 1

Question:

If $\displaystyle{\sf\:17\:\sin\:\theta\:=\:8}$, then find the value of $\displaystyle{\sf\:\dfrac{\tan\:\theta}{1\:+\:\tan^2\:\theta}}$

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{\tan\:\theta}{1\:+\:\tan^2\:\theta}\:=\:\dfrac{120}{289}}}}$

Step-by-step-explanation:

We have given that, $\displaystyle{\sf\:17\:\sin\:\theta\:=\:8}$

We have to find the value of $\displaystyle{\sf\:\dfrac{\tan\:\theta}{1\:+\:\tan^2\:\theta}}$

Now,

$\displaystyle{\sf\:17\:\sin\:\theta\:=\:8}$

$\displaystyle{\implies\sf\:\sin\:\theta\:=\:\dfrac{8}{17}}$

Now, we know that,

$\displaystyle{\pink{\sf\:\sin^2\:\theta\:+\:\cos^2\:\theta\:=\:1}\sf\:\:\:-\:-\:-\:[\:Trigonometric\:identity\:]}$

$\displaystyle{\implies\sf\:\cos^2\:\theta\:=\:1\:-\:\sin^2\:\theta}$

$\displaystyle{\implies\sf\:\cos\:\theta\:=\:\sqrt{1\:-\:\sin^2\:\theta}\:\:\:-\:-\:-\:[\:Taking\:square\:roots\:]}$

$\displaystyle{\implies\sf\:\cos\:\theta\:=\:\sqrt{1\:-\:\left(\:\dfrac{8}{17}\:\right)^2}}$

$\displaystyle{\implies\sf\:\cos\:\theta\:=\:\sqrt{1\:-\:\dfrac{8^2}{(\:17\:)^2}}\:\:\:\:-\:-\:-\:\left[\:\because\:\left(\:\dfrac{a}{b}\:\right)^m\:=\:\dfrac{a^m}{b^m}\:\right]}$

$\displaystyle{\implies\sf\:\cos\:\theta\:=\:\sqrt{1\:-\:\dfrac{64}{289}}}$

$\displaystyle{\implies\sf\:\cos\:\theta\:=\:\sqrt{\dfrac{289\:-\:64}{289}}}$

$\displaystyle{\implies\sf\:\cos\:\theta\:=\:\sqrt{\dfrac{225}{289}}}$

$\displaystyle{\implies\sf\:\cos\:\theta\:=\:\sqrt{\dfrac{15\:\times\:15}{17\:\times\:17}}}$

$\displaystyle{\implies\boxed{\red{\sf\:\cos\:\theta\:=\:\dfrac{15}{17}}}}$

Now, we know that,

$\displaystyle{\pink{\sf\:\tan\:\theta\:=\:\dfrac{\sin\:\theta}{\cos\:\theta}}\sf\:\:\:-\:-\:-\:[\:Trigonometric\:identity\:]}$

$\displaystyle{\implies\sf\:\tan\:\theta\:=\:\dfrac{\dfrac{8}{17}}{\dfrac{15}{17}}}$

$\displaystyle{\implies\sf\:\tan\:\theta\:=\:\dfrac{8}{\cancel{17}}\:\times\:\dfrac{\cancel{17}}{15}}$

$\displaystyle{\implies\boxed{\red{\sf\:\tan\:\theta\:=\:\dfrac{8}{15}}}}$

Now, we have to find the value of

$\displaystyle{\sf\:\dfrac{\tan\:\theta}{1\:+\:\tan^2\:\theta}}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{8}{15}}{1\:+\:\left(\:\dfrac{8}{15}\:\right)^2}}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{8}{15}}{1\:+\:\dfrac{64}{225}}}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{8}{15}}{\dfrac{225\:+\:64}{225}}}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{8}{15}}{\dfrac{289}{225}}}$

$\displaystyle{\implies\sf\:\dfrac{8}{\cancel{15}}\:\times\:\dfrac{\cancel{225}}{289}}$

$\displaystyle{\implies\sf\:\dfrac{8\:\times\:15}{289}}$

$\displaystyle{\implies\sf\:\dfrac{120}{289}}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\dfrac{\tan\:\theta}{1\:+\:\tan^2\:\theta}\:=\:\dfrac{120}{289}}}}}$