Question

The value of $\sin \left[ {{{\sin }^{ - 1}}\left( {\dfrac{1}{3}} \right) + {{\sec }^{ - 1}}\left( 3 \right)} \right] + \cos \left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( 2 \right)} \right]$ is
${\text{1) 1}}$
${\text{2) 2}}$
${\text{3) 3}}$
${\text{4) 4}}$

Answer 1

Hint : Convert ${\sec ^{ - 1}}x$ form of term into ${\cos ^{ - 1}}\left( {\dfrac{1}{x}} \right)$ form. Also convert ${\tan ^{ - 1}}x$ form of term into ${\cot ^{ - 1}}\left( {\dfrac{1}{x}} \right)$ form. After converting the equation we will use some results based on trigonometric functions to solve it further, then we put the values of $\sin \left( {\dfrac{\pi }{2}} \right)$ and $\cos \left( {\dfrac{\pi }{2}} \right)$ in the equation we obtain and can get our final result then.

Complete step-by-step answer :
The trigonometric functions are based on a very simple fact that we have to eliminate the additional information that is provided in the question by using basic trigonometric ideas and formulas. In this question we have to concentrate on two basic ideas that are properties of inverse trigonometric functions and their formulas. We know that inverse trigonometric functions can eliminate the basic trigonometric function if they are matched. But in this question in the parentheses two different functions are given so our basic aim is to make them the same and then apply formulas on them and by using basic steps we have to eliminate the inverse and simple trigonometric function. Now let’s start the solution,
The given equation is $\sin \left[ {{{\sin }^{ - 1}}\left( {\dfrac{1}{3}} \right) + {{\sec }^{ - 1}}\left( 3 \right)} \right] + \cos \left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( 2 \right)} \right]$ ------- $({\text{i}})$
When $\left| x \right| \geqslant 1$ then , ${\sec ^{ - 1}}x = {\cos ^{ - 1}}\left( {\dfrac{1}{x}} \right)$ is valid for values in the domain of ${\sec ^{ - 1}}x$ . Also , ${\tan ^{ - 1}}x = {\cot ^{ - 1}}\left( {\dfrac{1}{x}} \right)$ is valid for $x > 0$ . Therefore we can write ${\sec ^{ - 1}}\left( 3 \right)$ as ${\cos ^{ - 1}}\left( {\dfrac{1}{3}} \right)$ and ${\tan ^{ - 1}}\left( 2 \right)$ as ${\cot ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ . So, equation $({\text{i}})$ becomes
$\sin \left[ {{{\sin }^{ - 1}}\left( {\dfrac{1}{3}} \right) + {{\cos }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right] + \cos \left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\cot }^{ - 1}}\left( {\dfrac{1}{2}} \right)} \right]$ --------- $({\text{ii}})$
Because , ${\sin ^{ - 1}}\left( {\dfrac{1}{x}} \right) + {\cos ^{ - 1}}\left( {\dfrac{1}{x}} \right) = \dfrac{\pi }{2}$ and ${\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right) + {\cot ^{ - 1}}\left( {\dfrac{1}{x}} \right) = \dfrac{\pi }{2}$ . Therefore equation $({\text{ii}})$ becomes
$({\text{ii}}) \Rightarrow \sin \left( {\dfrac{\pi }{2}} \right) + \cos \left( {\dfrac{\pi }{2}} \right)$
The value of $\sin \left( {\dfrac{\pi }{2}} \right)$ is $1$ and the value of $\cos \left( {\dfrac{\pi }{2}} \right)$ is $0$ .
$\Rightarrow 1 + 0$
$\Rightarrow 1$
Hence, the correct option is ${\text{1) 1}}$ .
So, the correct answer is “Option B”.

Note : Keep in mind that ${\sec ^{ - 1}}x = {\cos ^{ - 1}}\left( {\dfrac{1}{x}} \right)$ and ${\tan ^{ - 1}}x = {\cot ^{ - 1}}\left( {\dfrac{1}{x}} \right)$ . Learning the formulas by heart will help a lot in solving various problems of trigonometry . Once you are clear with the formulas, try to do more and more problems for practice and to improve more. The symbol of ${\sin ^{ - 1}}x$ should not be confused with ${\left( {\sin x} \right)^{ - 1}}$ . Formulas play a very important role in trigonometry so try to remember all trigonometric formulas and values and keep practicing which helps in problem solving and remembering formulas.