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Question: If \[\tan ({\cos ^{ - 1}}x) = \sin ({\cot ^{ - 1}}\dfrac{1}{2})\], then find the value of \[x\]. A...

If tan(cos1x)=sin(cot112)\tan ({\cos ^{ - 1}}x) = \sin ({\cot ^{ - 1}}\dfrac{1}{2}), then find the value of xx.
A) x=53x = - \dfrac{{\sqrt 5 }}{3}
B) x=+53x = + \dfrac{{\sqrt 5 }}{3}
C) x=23x = - \dfrac{2}{3}
D) x=+23x = + \dfrac{2}{3}

Explanation

Solution

Hint: In the given question, we have to apply trigonometric identities to solve the question. We will have to find the hypotenuse of the triangle to arrive at the value of sinθ\sin \theta and then use trigonometric ratios to solve tanθ\tan \theta .

Complete step by step solution:

Let us solve the equation as follows:

First let us consider RHS term sin(cot112)\sin ({\cot ^{ - 1}}\dfrac{1}{2})

Using the trigonometric identity cot1θ=tan11θ{\cot ^{ - 1}}\theta = {\tan ^{ - 1}}\dfrac{1}{\theta }, we get,

=sin(tan12) = \sin ({\tan ^{ - 1}}2)

Using the formula, tanθ=OppositesideAdjacentside\tan \theta = \dfrac{{Opposite\,side\,}}{{Adjacent\,side\,}}, we can draw following diagram: (θ\theta is the angle ACB.)

Since tan1=2{\tan ^{ - 1}} = 2, we get

AB=2AB = 2and

BC=1BC = 1

Using Pythagoras Theorem (which states that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse), we get,

AC=AB2+BC2AC = \sqrt {A{B^2} + B{C^2}}

AC=(2)2+12AC = \sqrt {{{(2)}^2} + {1^2}}

AC=4+1\Rightarrow AC = \sqrt {4 + 1}

AC=5\Rightarrow AC = \sqrt 5

Therefore, we can get value of sinθ\sin \theta ,

sinθ=ABAC=25\sin \theta = \dfrac{{AB}}{{AC}} = \dfrac{2}{{\sqrt 5 }}

Hence RHS= 25\dfrac{2}{{\sqrt 5 }}

Now we can proceed to LHS side of equation tan(cos1x)\tan ({\cos ^{ - 1}}x):

Using the THS result, we will get,

tan(cos1x)=25\tan ({\cos ^{ - 1}}x) = \dfrac{2}{{\sqrt 5 }}

We can write the equation as follows:

cos1x=tan125{\cos ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{2}{{\sqrt 5 }}

Let tan125=t{\tan ^{ - 1}}\dfrac{2}{{\sqrt 5 }} = t

Therefore, we can say that,

25=tant\dfrac{2}{{\sqrt 5 }} = \tan t

Now let us draw the diagram again as follows:

Here tant=ABBC=25\tan t = \dfrac{{A'B'}}{{B'C'}} = \dfrac{2}{{\sqrt 5 }}

Using Pythagoras Theorem, we get,

AC=AB2+BC2A'C' = \sqrt {A'B{'^2} + B'C{'^2}}

AC=(2)2+(5)2A'C' = \sqrt {{{(2)}^2} + {{(\sqrt 5 )}^2}}

AC=4+5A'C' = \sqrt {4 + 5}

AC=9A'C' = \sqrt 9

AC=3A'C' = 3

Therefore, we can get the value of cost=BCAC\cos t = \dfrac{{B'C'}}{{A'C'}}

cost=53\cos t = \dfrac{{\sqrt 5 }}{3}

Since tan125=t{\tan ^{ - 1}}\dfrac{2}{{\sqrt 5 }} = t, we get,

cos1x=t{\cos ^{ - 1}}x = t

Using the property cos1A=B{\cos ^{ - 1}}A = Bso A=cosBA = \cos B, we will get:

x=costx = \cos t

x=53x = \dfrac{{\sqrt 5 }}{3}

Hence, option (B) x=+53x = + \dfrac{{\sqrt 5 }}{3} is the correct answer.

Note:

  1. cot1θ=tan11θ{\cot ^{ - 1}}\theta = {\tan ^{ - 1}}\dfrac{1}{\theta } is proved as below:

Let cot1(x)=θ{\cot ^{ - 1}}(x) = \theta

So, we get x=cotθx = \cot \theta

We know that cotangent is the inverse of tangent. So, we get,

1x=tanθ\dfrac{1}{x} = \tan \theta

Therefore, we can say that,

θ=tan1(1x)=cot1(x)\theta = {\tan ^{ - 1}}(\dfrac{1}{x}) = {\cot ^{ - 1}}(x)

  1. Meaning of the terms used to find the missing value in diagram are clarified below:

Sine: The ratio of side opposite to given angle and hypotenuse is called sine. It is denoted as sinθ\sin \theta .

sinθ=SideoppositetogivenangleHypotenuse\sin \theta = \dfrac{{Side\,opposite\,to\,given\,angle}}{{Hypotenuse}}

In the given sum, we will get sinθ=ABAC\sin \theta = \dfrac{{AB}}{{AC}}

Tangent: The ratio of side opposite to given angle and its adjacent side is called tangent. It is denoted as tanθ\tan \theta .

tanθ=SideoppositetogivenangleSideadjacenttogivenangle\tan \theta = \dfrac{{Side\,opposite\,to\,given\,angle}}{{Side\,adjacent\,to\,given\,angle}}

In the given sum, we will get tanθ=ABBC\tan \theta = \dfrac{{A'B'}}{{B'C'}}