Question

if k=2(sin15/cos75+cot78/tan12)-1 then k=​

Answer 1

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Answer 2

ANSWER:

Given:

• k = 2(sin 15/cos 75 + cot 78/tan 12) - 1

To Find:

• Value of k

Solution:

We are given that,

$\implies k=2\left(\dfrac{\sin15}{\cos75}+\dfrac{\cot78}{\tan12}\right)-1$

We know that,

$\hookrightarrow \sin\theta=\cos(90-\theta)$

And,

$\hookrightarrow \cot\theta=\tan(90-\theta)$

So,

$\implies k=2\left(\dfrac{\sin15}{\cos75}+\dfrac{\cot78}{\tan12}\right)-1$

$\implies k=2\left(\dfrac{\cos(90-15)}{\cos75}+\dfrac{\tan(90-78)}{\tan12}\right)-1$

$\implies k=2\left(\dfrac{\cos15}{\cos75}+\dfrac{\tan12}{\tan12}\right)-1$

$\implies k=2(1+1)-1$

$\implies k=2(2)-1$

$\implies k=4-1$

$\implies\bf k=3$

Formula Used:

• $\hookrightarrow \sin\theta=\cos(90-\theta)$
• $\hookrightarrow \cot\theta=\tan(90-\theta)$

Learn More:

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

$\begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}$