Solveeit Logo
Login

Question

Question: How to solve for \( \tan (\dfrac{x}{2}) \) if \( \tan x = ( - \dfrac{5}{{12}}) \) ?...

How to solve for tan⁡(x2)\tan (\dfrac{x}{2}) if tan⁡x=(−512)\tan x = ( - \dfrac{5}{{12}}) ?

Explanation

Solution

Hint : In this question, we are given the value of the tangent of an angle x and we have to find the tangent of half of the angle x. We know that the tangent function is the ratio of the sine function and cosine function, so the tangent of half of x can be written as the ratio of the sine of half of x and the cosine of half of x, finding the value of sine and cosine by the use of identities, we can eventually find out the required value.

Complete step-by-step answer :
We are given that tan⁡x=(−512)\tan x = ( - \dfrac{5}{{12}}) , we know that tangent is negative in the second and fourth quadrant and we know that –
sec⁡2x−tan⁡2x=1 ⇒sec⁡2x=1+(−512)2 ⇒sec⁡2x=144+25144 ⇒sec⁡x=169144 ⇒sec⁡x=±1312 ⇒cos⁡x=1sec⁡x=±1213   {\sec ^2}x - {\tan ^2}x = 1 \\\ \Rightarrow {\sec ^2}x = 1 + {(\dfrac{{ - 5}}{{12}})^2} \\\ \Rightarrow {\sec ^2}x = \dfrac{{144 + 25}}{{144}} \\\ \Rightarrow \sec x = \sqrt {\dfrac{{169}}{{144}}} \\\ \Rightarrow \sec x = \pm \dfrac{{13}}{{12}} \\\ \Rightarrow \cos x = \dfrac{1}{{\sec x}} = \pm \dfrac{{12}}{{13}} \;
Now, cosine is negative in the second quadrant and positive in the fourth quadrant, so we will take both these values.
We know that
sin⁡2x+cos⁡2x=1 ⇒sin⁡2x=1−cos⁡2x ⇒sin⁡x=1−144169=25169=±513   {\sin ^2}x + {\cos ^2}x = 1 \\\ \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x \\\ \Rightarrow \sin x = \sqrt {1 - \dfrac{{144}}{{169}}} = \sqrt {\dfrac{{25}}{{169}}} = \pm \dfrac{5}{{13}} \;
Cosine is negative in the second quadrant but sine is positive, so when cos⁡x=−1213, sin⁡x=513\cos x = - \dfrac{{12}}{{13}},\,\sin x = \dfrac{5}{{13}} and in the fourth quadrant cosine is positive but sine is negative, so when cos⁡x=1213, sin⁡x=−513\cos x = \dfrac{{12}}{{13}},\,\sin x = - \dfrac{5}{{13}}
Now, tan⁡x2=sin⁡x2cos⁡x2\tan \dfrac{x}{2} = \dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}
Multiplying 2cos⁡x22\cos \dfrac{x}{2} with both the numerator and the denominator –
⇒tan⁡x2=2sin⁡x2cos⁡x22cos⁡2x2=2sin⁡x2cos⁡x22cos⁡2x2−1+1=sin⁡xcos⁡x+1\Rightarrow \tan \dfrac{x}{2} = \dfrac{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}{{2{{\cos }^2}\dfrac{x}{2}}} = \dfrac{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}{{2{{\cos }^2}\dfrac{x}{2} - 1 + 1}} = \dfrac{{\sin x}}{{\cos x + 1}}
Putting the value of sine and cosine in the above-obtained formula, we get –
At cos⁡x=−1213, sin⁡x=513\cos x = \dfrac{{ - 12}}{{13}},\,\sin x = \dfrac{5}{{13}}
tan⁡x2=5131+(−1213)=513×131 ⇒tan⁡x2=5   \tan \dfrac{x}{2} = \dfrac{{\dfrac{5}{{13}}}}{{1 + (\dfrac{{ - 12}}{{13}})}} = \dfrac{5}{{13}} \times \dfrac{{13}}{1} \\\ \Rightarrow \tan \dfrac{x}{2} = 5 \;
At cos⁡x=1213, sin⁡x=−513\cos x = \dfrac{{12}}{{13}},\,\sin x = \dfrac{{ - 5}}{{13}}
tan⁡x2=−5131+1213=−513×1325 ⇒tan⁡x2=−15   \tan \dfrac{x}{2} = \dfrac{{\dfrac{{ - 5}}{{13}}}}{{1 + \dfrac{{12}}{{13}}}} = \dfrac{{ - 5}}{{13}} \times \dfrac{{13}}{{25}} \\\ \Rightarrow \tan \dfrac{x}{2} = - \dfrac{1}{5} \;
Hence, if tan⁡x=−512\tan x = \dfrac{{ - 5}}{{12}} , tan⁡x2=5 or tan⁡x2=−15\tan \dfrac{x}{2} = 5\,or\,\tan \dfrac{x}{2} = - \dfrac{1}{5}
So, the correct answer is “ tan⁡x2=5 or tan⁡x2=−15\tan \dfrac{x}{2} = 5\,or\,\tan \dfrac{x}{2} = - \dfrac{1}{5} ”.

Note : There are several formulas for finding the sine and cosine of the half of a given angle but there is no such formula for finding the tangent of the half of the given angle, so by using the knowledge of identities or of trigonometric ratios, we can find out the value of sine and cosine of angle x and then plug in those values in the formula of the tangent function.