Question

Solve \[{{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})=\]
A.\[\dfrac{\pi }{3}\]
B.\[\dfrac{\pi }{4}\]
C.\[\dfrac{\pi }{6}\]
D.\[\dfrac{\pi }{2}\]

Answer 1

Hint: An equation which involves trigonometric ratio of any angle is said to be a trigonometric identity if it is satisfied for all values for which the given trigonometric ratios are defined. Trigonometric functions are periodic functions and all trigonometric functions are not bijections. Consequently their inverse does not exist. If no branch of an inverse trigonometric function is given, then it means that the principal value branch of the function. An inverse function reverses the direction of the original function.

Complete step-by-step answer:
Let us assume that \[x=\cos e{{c}^{-1}}(\sqrt{5})\]
By solving we get
\[\Rightarrow \cos ecx=\sqrt{5}\]
Using the trigonometric identity \[\cos e{{c}^{2}}\theta =1+{{\cot }^{2}}\theta \]
Rearranging the identity we get
\[{{\cot }^{2}}x=\cos e{{c}^{2}}x-1\]
Substituting the value we get
\[{{\cot }^{2}}x={{(\sqrt{5})}^{2}}-1\]
Further simplifying we get
\[{{\cot }^{2}}x=5-1\]
On solving we get
\[{{\cot }^{2}}x=4\]
Taking square root on both sides we get
\[\cot x=2\]
Further rearranging we get
\[x={{\cot }^{-1}}(2)\]
Equating the values we get
\[\cos e{{c}^{-1}}(\sqrt{5})={{\cot }^{-1}}(2)\]
So we can write the equation as
\[{{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})={{\cot }^{-1}}(3)+{{\cot }^{-1}}(2)\]
The above equation can be written as
\[{{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})={{\tan }^{-1}}\dfrac{1}{3}+{{\tan }^{-1}}\dfrac{1}{2}\]
Now we will use the trigonometric formula given as
\[{{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)\]
Using this trigonometric formula we get
\[{{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})={{\tan }^{-1}}\left( \dfrac{\dfrac{1}{3}+\dfrac{1}{2}}{1-\dfrac{1}{3}\times \dfrac{1}{2}} \right)\]
Further simplifying we get
\[{{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})={{\tan }^{-1}}\left( \dfrac{5}{5} \right)\]
On solving we get
\[{{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})={{\tan }^{-1}}(1)\]
Using trigonometric ratios and angles concept to solve further we get
\[{{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})={{\tan }^{-1}}(\tan \dfrac{\pi }{4})\]
As we know that
\[{{\tan }^{-1}}(\tan \theta )=\theta \]
Applying this we get
\[{{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})=\dfrac{\pi }{4}\]
Therefore, option \[B\] is the correct answer.
So, the correct answer is “Option B”.

Note: Before solving the trigonometric problems, one must be familiar with the trigonometric ratios, trigonometric identities, inverse trigonometric functions and trigonometric applications. The word ‘trigonometry’ is derived from the Greek words ‘tri’ which means three, ‘gon’ (means sides) and ‘metron’ (means measure). Some ratios of the sides with respect to its acute angles, called trigonometric ratios of the angle.