Question

Find the value of $\tan \left( {3{{\tan }^{ - 1}}3} \right) + \cos \left( {3{{\cos }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right) + 1$

(A) 11

(B) $\dfrac{9}{{13}}$

(C) $\dfrac{4}{{27}}$

(D) $\dfrac{{295}}{{351}}$

Answer 1

Hint: In this question, you can simplify using the formulas of $\tan 3A$ and $\cos 3A$. Calculate $\tan \left( {3{{\tan }^{ - 1}}3} \right)$ and $\cos \left( {3{{\cos }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)$ separately and put the calculated values in the question and then simplify to get the result.

Formula used:

Here we use the formula according to the question, we need $\tan 3A = \dfrac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}$ and $\cos 3A = 4{\cos ^3}A - 3\cos A$

Complete step-by-step answer:

We are given that, $\tan \left( {3{{\tan }^{ - 1}}3} \right) + \cos \left( {3{{\cos }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right) + 1$

Now, we are using the formula to calculate:

$\tan 3A = \dfrac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}$ here in this part of the question that is $\tan \left( {3{{\tan }^{ - 1}}3} \right)$ where $A = {\tan ^{ - 1}}3$

By substituting the value of A in the formula we get, $ = \dfrac{{3\tan \left( {{{\tan }^{ - 1}}3} \right) - {{\tan }^3}\left( {{{\tan }^{ - 1}}3} \right)}}{{1 - 3{{\tan }^2}\left( {{{\tan }^{ - 1}}3} \right)}}$ and as we know $\tan ({\tan ^{ - 1}}A) = A$ .

Therefore, = $\dfrac{{3 * 3 - {{\left( 3 \right)}^3}}}{{1 - {{\left( 3 \right)}^3}}}$

On simplifying we get,

$ = \dfrac{{9 - 27}}{{1 - 27}}$

$ = \dfrac{{ - 18}}{{ - 26}}$

Cancelling negative sign and dividing by 2 from both numerator and denominator.

We get, $ = \dfrac{9}{{13}}$

Hence, $\tan \left( {3{{\tan }^{ - 1}}3} \right)$$ = \dfrac{9}{{13}}$

Similarly we can find out for $\cos \left( {3{{\cos }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)$ by using the formula $\cos 3A = 4{\cos ^3}A - 3\cos A$,

Here, $A = {\cos ^{ - 1}}\left( {\dfrac{1}{3}} \right)$

By substituting the value of A in the formula we get,

$ = 4{\cos ^3}\left( {{{\cos }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right) - 3\cos \left( {{{\cos }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)$

As, we know $\cos ({\cos ^{ - 1}}A) = A$

Therefore, $ \Rightarrow \dfrac{4}{{{{\left( 3 \right)}^3}}} - 1$

On simplifying we get,

$ \Rightarrow \dfrac{4}{{27}} - 1$

By taking L.C.M and simplifying we get,

$ \Rightarrow \dfrac{{4 - 27}}{{27}}$

$ \Rightarrow \dfrac{{ - 23}}{{27}}$

Hence, $\cos \left( {3{{\cos }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)$ = $\dfrac{{ - 23}}{{27}}$

Putting all the values of $\tan \left( {3{{\tan }^{ - 1}}3} \right)$ and $\cos \left( {3{{\cos }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)$ on given equation:

We get,

= $\dfrac{9}{{13}} - \dfrac{{23}}{{27}} + 1$

By taking L.C.M and simplifying we get,

= $\dfrac{{243 - 299 + 351}}{{351}}$

= $\dfrac{{594 - 299}}{{351}}$

= $\dfrac{{295}}{{351}}$

 Hence, $\tan \left( {3{{\tan }^{ - 1}}3} \right) + \cos \left( {3{{\cos }^{ - 1}}3} \right) + 1$ = $\dfrac{{295}}{{351}}$

So, option (D) is correct.

Additional Information: Trigonometric identities are the formulas which include trigonometric functions. These trigonometric formulas help us to solve the question in a simpler way. There are a very large number of identities that can be used in various fields and have a lot of applications.

Note: These types of questions are done with the help of trigonometric formulas. Use the formulas according to the ask of the question. And simplify the question to get the desired result by putting the values.