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Question: How do you evaluate \(\tan \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right)\) witho...

How do you evaluate tan⁡(cos⁡−1(−23))\tan \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right) without a calculator?

Explanation

Solution

This problem deals with applying the basic and important trigonometric identities. We are given a tangent trigonometric expression inside of which there is an inverse of cosine trigonometric expression of a particular value. So in order to proceed to get the exact value of the expression, first we need to assign the given inverse cosine trigonometric value to a variable, and then solve.

Complete step-by-step solution:
Given the expression of trigonometric and inverse trigonometric ratio which is tan⁡(cos⁡−1(−23))\tan \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right).
Consider the given expression, as given below:
⇒tan⁡(cos⁡−1(−23))\Rightarrow \tan \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right)
Now consider the inside of tangent value which is cos⁡−1(−23){\cos ^{ - 1}}\left( { - \dfrac{2}{3}} \right), as given below:
Let the expression of cos⁡−1(−23){\cos ^{ - 1}}\left( { - \dfrac{2}{3}} \right) is equal to α\alpha , which is mathematically expressed below:
⇒α=cos⁡−1(−23)\Rightarrow \alpha = {\cos ^{ - 1}}\left( { - \dfrac{2}{3}} \right)
Now take inverse cosine trigonometric function on both the sides of the above equation, as shown below:
⇒cos⁡α=cos⁡(cos⁡−1(−23))\Rightarrow \cos \alpha = \cos \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right)
Here on the right hand side of the above equation, cosine and inverse cosine trigonometric function gets cancelled, as shown below:
⇒cos⁡α=−23\Rightarrow \cos \alpha = - \dfrac{2}{3}
Now as we considered α=cos⁡−1(−23)\alpha = {\cos ^{ - 1}}\left( { - \dfrac{2}{3}} \right), hence the expression tan⁡(cos⁡−1(−23))\tan \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right) becomes as shown below:
⇒tan⁡(cos⁡−1(−23))=tan⁡α\Rightarrow \tan \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right) = \tan \alpha
So if we find the value of tan⁡α\tan \alpha , then it is the same as finding the value oftan⁡(cos⁡−1(−23))\tan \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right).
Hence finding the value of tan⁡α\tan \alpha .
But we know the value of cos⁡α\cos \alpha , which is equal to −23 - \dfrac{2}{3}.
Hence to find the value of tan⁡α\tan \alpha , we can express tan⁡α\tan \alpha in terms of cos⁡α\cos \alpha , and then can get the value of tan⁡α\tan \alpha .
So expressing tan⁡α\tan \alpha in terms of cos⁡α\cos \alpha , as given below:
⇒tan⁡α=sin⁡αcos⁡α\Rightarrow \tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}
We know that from the basic trigonometric identity sin⁡2α+cos⁡2α=1{\sin ^2}\alpha + {\cos ^2}\alpha = 1, from here the value of sin⁡α\sin \alpha can be written as:
⇒sin⁡2α=1−cos⁡2α\Rightarrow {\sin ^2}\alpha = 1 - {\cos ^2}\alpha
∴sin⁡α=1−cos⁡2α\therefore \sin \alpha = \sqrt {1 - {{\cos }^2}\alpha }
Now substituting the above expression in the expression of tan⁡α\tan \alpha , as given below:
⇒tan⁡α=1−cos⁡2αcos⁡α\Rightarrow \tan \alpha = \dfrac{{\sqrt {1 - {{\cos }^2}\alpha } }}{{\cos \alpha }}
We obtained that the value of cos⁡α=−23\cos \alpha = - \dfrac{2}{3}, hence substituting it in the above expression, as shown:
⇒tan⁡α=1−(−23)2(−23)\Rightarrow \tan \alpha = \dfrac{{\sqrt {1 - {{\left( { - \dfrac{2}{3}} \right)}^2}} }}{{\left( { - \dfrac{2}{3}} \right)}}
Simplifying the above expression, as given below:
⇒tan⁡α=−321−(49)\Rightarrow \tan \alpha = - \dfrac{3}{2}\sqrt {1 - \left( {\dfrac{4}{9}} \right)}
⇒tan⁡α=−3259\Rightarrow \tan \alpha = - \dfrac{3}{2}\sqrt {\dfrac{5}{9}}
As we know that the value of under root of 9 is 3, 9=3\sqrt 9 = 3, as shown:
⇒tan⁡α=−32×35\Rightarrow \tan \alpha = - \dfrac{3}{{2 \times 3}}\sqrt 5
⇒tan⁡α=−52\Rightarrow \tan \alpha = - \dfrac{{\sqrt 5 }}{2}

Hence the value of tan⁡(cos⁡−1(−23))=−52\tan \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right) = - \dfrac{{\sqrt 5 }}{2}.

Note: Please note that while solving any trigonometric based problems, we need to be through with all the important and basic trigonometric identities, few are given below:
⇒sin⁡2α+cos⁡2α=1\Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = 1
From which we can obtain sin⁡α=1−cos⁡2α\sin \alpha = \sqrt {1 - {{\cos }^2}\alpha }
⇒sec⁡2α−tan⁡2α=1\Rightarrow {\sec ^2}\alpha - {\tan ^2}\alpha = 1
From which we can obtain tan⁡α=sec⁡2α−1\tan \alpha = \sqrt {{{\sec }^2}\alpha - 1}
⇒cos⁡ec2α−cot⁡2α=1\Rightarrow \cos e{c^2}\alpha - {\cot ^2}\alpha = 1
From which we can obtain cot⁡α=cos⁡ec2α−1\cot \alpha = \sqrt {\cos e{c^2}\alpha - 1}