Question

If 5sec theta-12cosec theta=0 find the values of sec theta,cos theta and sin theta

Answer 1

Answer:

$\large{\boxed{\red{\sf\:\sec\:\theta\:=\:\dfrac{13}{5}\:,\:\cos\:\theta\:=\:\dfrac{5}{13}\:,\:\sin\:\theta\:=\:\dfrac{12}{13}}}}$

Step-by-step-explanation:

We have given that,

$\sf\:5\:\sec\:\theta\:-\:12\:\csc\:\theta\:=\:0\\\\\\\implies\sf\:5\:\sec\:\theta\:=\:12\:\csc\:\theta\\\\\\\implies\sf\:\dfrac{\sec\:\theta}{\csc\:\theta}\:=\:\dfrac{12}{5}\\\\\\\implies\sf\:\sec\:\theta\:\div\:\csc\:\theta\:=\:\dfrac{12}{5}\\\\\\\implies\sf\:\dfrac{1}{\cos\:\theta}\:\div\:\dfrac{1}{\sin\:\theta}\:=\:\dfrac{12}{5}\:\:-\:-\:[\:\sec\:\theta\:=\:\dfrac{1}{\cos\:\theta}\:,\:\csc\:\theta\:=\:\dfrac{1}{\sin\:\theta}\:]\\\\\\\implies\sf\:\dfrac{1}{\cos\:\theta}\:\times\:\dfrac{\sin\:\theta}{1}\:=\:\dfrac{12}{5}\\\\\\\implies\sf\:\dfrac{\sin\:\theta}{\cos\:\theta}\:=\:\dfrac{12}{5}\\\\\\\implies\sf\:\tan\:\theta\:=\:\dfrac{12}{5}$

Now,

$\pink{\sf\:1\:+\:\tan^2\:theta\:=\:\sec^2\:\theta}\\\\\\\implies\sf\:1\:+\:\bigg(\:\dfrac{12}{5}\:\bigg)^2\:=\:\sec^2\:\theta\\\\\\\implies\sf\:\sec^2\:\theta\:=\:1\:+\:\dfrac{144}{25}\\\\\\\implies\sf\:\sec^2\:\theta\:=\:\dfrac{25\:+\:144}{25}\\\\\\\implies\sf\:\sec^2\:\theta\:=\:\dfrac{169}{25}\\\\\\\implies\boxed{\red{\sf\:\sec^2\:\theta\:=\:\dfrac{13}{5}}}\sf\:\:\:-\:-\:[\:Taking\:square\:roots\:]$

Now,

$\sf\:\pink{\sf\:\cos\:\theta\:=\:\dfrac{1}{\sec\:\theta}}\\\\\\\implies\sf\:\cos\:\theta\:=\:\dfrac{1}{\bigg(\:\dfrac{13}{5}\:\bigg)}\\\\\\\implies\boxed{\red{\sf\:\cos\:\theta\:=\:\dfrac{5}{13}}}$

Now,

$\pink{\sf\:\sin^2\:\theta\:+\:\cos^2\:\theta\:=\:1}\\\\\\\implies\sf\:\sin^2\:\theta\:+\:\bigg(\:\dfrac{5}{13}\:\bigg)^2\:=\:1\\\\\\\implies\sf\:\sin^2\:\theta\:+\:\dfrac{25}{169}\:=\:1\\\\\\\implies\sf\:\sin^2\:\theta\:=\:1\:-\:\dfrac{25}{169}\\\\\\\implies\sf\:\sin^2\:\theta\:=\:\dfrac{169\:-\:25}{169}\\\\\\\implies\sf\:\sin^2\:\theta\:=\:\dfrac{144}{169}\\\\\\\implies\boxed{\red{\sf\:\sin^2\:\theta\:=\:\dfrac{12}{13}}}\sf\:\:\:-\:-\:[\:Taking\:square\:roots\:]$

$\large{\boxed{\red{\sf\:\sec\:\theta\:=\:\dfrac{13}{5}\:,\:\cos\:\theta\:=\:\dfrac{5}{13}\:,\:\sin\:\theta\:=\:\dfrac{12}{13}}}}$