Question

The value of $\cot \left[ {\cos e{c^{ - 1}}\left( {\dfrac{5}{3}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right]$ is
A. $\dfrac{3}{{17}}$
B. $\dfrac{4}{{17}}$
C. $\dfrac{5}{{17}}$
D. $\dfrac{6}{{17}}$

Answer 1

Use various trigonometric identities. Use Pythagoras Theorem (Pythagorean Theorem). Trigonometric and their inverse trigonometric parts get cancelled or compensated.

Complete step by step answer:
We need to find the value of $\cot \left[ {\cos e{c^{ - 1}}\left( {\dfrac{5}{3}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right]$
Using the property that $\cos e{c^{ - 1}}\left( {\dfrac{5}{3}} \right) = {\sin ^{ - 1}}\left( {\dfrac{3}{5}} \right)$
Also by using Pythagoras Theorem which is also sometimes called the Pythagorean Theorem. (Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. The sides of this triangle have been named as Perpendicular, Base and Hypotenuse. Here, the hypotenuse is the longest side, as it is opposite to the angle 90°. The sides of a right triangle (say a, b and c) which have positive integer values, when squared, are put into an equation, also called a Pythagorean triple.)
Consider a right angled triangle :
Where “a” is the perpendicular,
“b” is the base,
“c” is the hypotenuse.
According to the definition, the Pythagoras Theorem formula is given as: ${\left( {Hypotenuse} \right)^2} = {\left( {Perpendicular} \right)^2} + {\left( {Base} \right)^2}$
That is ${\left( c \right)^2} = {\left( a \right)^2} + {\left( b \right)^2}$
Trigonometric functions in terms of sides of a right angled triangle:
$\sin \theta = \dfrac{a}{c}$
$\cos \theta = \dfrac{b}{c}$
$\tan \theta = \dfrac{a}{b}$
We know that
${\sin ^{ - 1}}\left( {\dfrac{3}{5}} \right) = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$
Therefore $\cot \left[ {\cos e{c^{ - 1}}\left( {\dfrac{5}{3}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right]$ reduces to $\cot \left[ {{{\tan }^{ - 1}}\left( {\dfrac{3}{4}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right]$
Now by using the property of inverse trigonometric functions this further changes to
$\cot \left[ {{{\tan }^{ - 1}}\dfrac{{\dfrac{3}{4} + \dfrac{2}{3}}}{{1 - \dfrac{3}{4}*\dfrac{2}{3}}}} \right]$
On simplifying this we get
$= \cot \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{\dfrac{{17}}{{12}}}}{{\dfrac{1}{2}}}} \right)} \right]$
On simplifying this we get
$= \cot \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{17}}{6}} \right)} \right]$
On changing ${\tan ^{ - 1}}$ to ${\cot ^{ - 1}}$ we get
$= \cot \left[ {{{\cot }^{ - 1}}\left( {\dfrac{6}{{17}}} \right)} \right]$
Now cancelling the trigonometric and inverse trigonometric part we get
$= \dfrac{6}{{17}}$

So, the correct answer is “Option D”.

Note: Use various trigonometric identities. Always try to make the trigonometric and its inverse part so as to ease the calculations. Trigonometric and their inverse trigonometric parts get cancelled or compensated.