# Find the value of the expression $\tan A+\tan \left( 60+A \right)-\tan \left( 60-A \right)$ is equal to

a) 3 tan 3A

b) tan 3A

c) cot 3A

d) sin 3A

Last updated date: 16th Mar 2023

•

Total views: 305.4k

•

Views today: 2.85k

Answer

Verified

305.4k+ views

Hint: Use trigonometric identities to simplify the relation. Convert given tan functions to sin and cos functions. Take any two tan functions to solve the given expression easily.

Trigonometric expression given here is;

$\tan A+\tan \left( 60+A \right)-\tan \left( 60-A \right)$

Let, we represent the above relation by ‘M’, hence;

$M=\tan A+\tan \left( 60+A \right)-\tan \left( 60-A \right)$……………….. (1)

Let us solve last two terms by converting tan to sin and cos using relation;

$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ …………………. (2)

Hence, equation (1), can be written as;

$M=\tan A+\dfrac{\sin \left( 60+A \right)}{\cos \left( 60+A \right)}-\dfrac{\sin \left( 60-A \right)}{\cos \left( 60-A \right)}$

Taking L.C.M in last two fractions, we get

$M=\tan A+\dfrac{\left( \sin \left( 60+A \right).\cos \left( 60-A \right)-\sin \left( 60-A \right).\cos \left( 60+A \right) \right)}{\cos \left( 60+A \right).\cos \left( 60-A \right)}$

Now, we know the trigonometric identities as;

$\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ or vice – versa can also be used.

And, we also know that

$\cos \left( A-B \right)\cos \left( A+B \right)={{\cos }^{2}}B-{{\sin }^{2}}A$

Using the above trigonometric identities in for simplifying the relation ‘M’ we get;

$\begin{align}

& M=\tan A+\dfrac{\sin {\left( 60+A \right)-\left( 60-A \right)}}{{{\cos }^{2}}A-{{\sin }^{2}}{60}} \\

& M=\tan A+\dfrac{\sin 2A}{{{\cos }^{2}}A-{{\sin }^{2}}60{}^\circ } \\

\end{align}$

Now, we know the value of \[sin\text{ }60{}^\circ \]i.e. $\dfrac{\sqrt{3}}{2}$. Putting value of ${{\sin }^{2}}60$, in above equation, we get;

$\begin{align}

& M=\tan A+\dfrac{\sin 2A}{{{\cos }^{2}}A-\dfrac{3}{4}} \\

& or \\

& M=\tan A+\dfrac{4\sin 2A}{4{{\cos }^{2}}A-3} \\

\end{align}$

Now, we can use trigonometric identity $\sin 2A=2\sin A\cos A\text{ and }\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ to simplify the above relation more, we get;

$M=\dfrac{\sin A}{\cos A}+\dfrac{8\sin A\cos A}{4{{\cos }^{2}}A-3}$

Now, taking ‘sinA’ common from both the terms and L.C.M as well, we get;

$\begin{align}

& M=\sin A\left( \dfrac{1}{\cos A}+\dfrac{8\cos A}{4{{\cos }^{2}}A-3} \right) \\

& or \\

& M=\sin A\left( \dfrac{4{{\cos }^{2}}A-3+8{{\cos }^{2}}A}{\cos A\left( 4{{\cos }^{2}}A-3 \right)} \right) \\

& M=\sin A\left( \dfrac{12{{\cos }^{2}}A-3}{\cos A\left( 4{{\cos }^{2}}A-3 \right)} \right) \\

\end{align}$

Now, using the relation ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\ or\ {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ to replace ${{\cos }^{2}}A\ \text{ by}\ 1-{{\sin }^{2}}A$, we get;

$\begin{align}

& M=\sin A\left( \dfrac{12\left( 1-{{\sin }^{2}}A \right)-3}{\cos A\left( 4{{\cos }^{2}}A-3 \right)} \right) \\

& M=\sin A\left( \dfrac{9-12{{\sin }^{2}}A}{\cos A\left( 4{{\cos }^{2}}A-3 \right)} \right) \\

\end{align}$

Taking ‘3’ common from numerator, we get;

$\begin{align}

& M=3\sin A\left( \dfrac{3-4{{\sin }^{2}}A}{\cos A\left( 4{{\cos }^{2}}A-3 \right)} \right) \\

& or \\

& M=3\sin A\left( \dfrac{3\sin A-4{{\sin }^{3}}A}{4{{\cos }^{3}}A-3\cos A} \right) \\

\end{align}$

AS, we know trigonometric identities for converting $3\theta \ to\ \theta $ with sin and cos terms as follows;

$\begin{align}

& \sin 3A=3\sin A-4{{\sin }^{3}}A \\

& \cos 3A=4{{\cos }^{3}}A-3\cos A \\

\end{align}$

Hence, above relation of M can be re-written as;

$M=3\dfrac{\sin 3A}{\cos 3A}=3\tan 3A$

Hence, the value of tan A + tan (60 + A) – tan (60 – A) is 3 tan 3A.

So, option A is the correct answer.

Note: One can get confused with the formulae of sin 3A and cos 3A as mentioned in the solution.

Another approach for this question would be that we can use trigonometry identities;

$\begin{align}

& \tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} \\

& and \\

& \tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B} \\

\end{align}$

to simplify tan (60 + A) and tan (60 – A). Then simplify the results found to get the answer. Applying trigonometric identities always makes solutions flexible and shorter which is the key point of the solution.

Trigonometric expression given here is;

$\tan A+\tan \left( 60+A \right)-\tan \left( 60-A \right)$

Let, we represent the above relation by ‘M’, hence;

$M=\tan A+\tan \left( 60+A \right)-\tan \left( 60-A \right)$……………….. (1)

Let us solve last two terms by converting tan to sin and cos using relation;

$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ …………………. (2)

Hence, equation (1), can be written as;

$M=\tan A+\dfrac{\sin \left( 60+A \right)}{\cos \left( 60+A \right)}-\dfrac{\sin \left( 60-A \right)}{\cos \left( 60-A \right)}$

Taking L.C.M in last two fractions, we get

$M=\tan A+\dfrac{\left( \sin \left( 60+A \right).\cos \left( 60-A \right)-\sin \left( 60-A \right).\cos \left( 60+A \right) \right)}{\cos \left( 60+A \right).\cos \left( 60-A \right)}$

Now, we know the trigonometric identities as;

$\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ or vice – versa can also be used.

And, we also know that

$\cos \left( A-B \right)\cos \left( A+B \right)={{\cos }^{2}}B-{{\sin }^{2}}A$

Using the above trigonometric identities in for simplifying the relation ‘M’ we get;

$\begin{align}

& M=\tan A+\dfrac{\sin {\left( 60+A \right)-\left( 60-A \right)}}{{{\cos }^{2}}A-{{\sin }^{2}}{60}} \\

& M=\tan A+\dfrac{\sin 2A}{{{\cos }^{2}}A-{{\sin }^{2}}60{}^\circ } \\

\end{align}$

Now, we know the value of \[sin\text{ }60{}^\circ \]i.e. $\dfrac{\sqrt{3}}{2}$. Putting value of ${{\sin }^{2}}60$, in above equation, we get;

$\begin{align}

& M=\tan A+\dfrac{\sin 2A}{{{\cos }^{2}}A-\dfrac{3}{4}} \\

& or \\

& M=\tan A+\dfrac{4\sin 2A}{4{{\cos }^{2}}A-3} \\

\end{align}$

Now, we can use trigonometric identity $\sin 2A=2\sin A\cos A\text{ and }\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ to simplify the above relation more, we get;

$M=\dfrac{\sin A}{\cos A}+\dfrac{8\sin A\cos A}{4{{\cos }^{2}}A-3}$

Now, taking ‘sinA’ common from both the terms and L.C.M as well, we get;

$\begin{align}

& M=\sin A\left( \dfrac{1}{\cos A}+\dfrac{8\cos A}{4{{\cos }^{2}}A-3} \right) \\

& or \\

& M=\sin A\left( \dfrac{4{{\cos }^{2}}A-3+8{{\cos }^{2}}A}{\cos A\left( 4{{\cos }^{2}}A-3 \right)} \right) \\

& M=\sin A\left( \dfrac{12{{\cos }^{2}}A-3}{\cos A\left( 4{{\cos }^{2}}A-3 \right)} \right) \\

\end{align}$

Now, using the relation ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\ or\ {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ to replace ${{\cos }^{2}}A\ \text{ by}\ 1-{{\sin }^{2}}A$, we get;

$\begin{align}

& M=\sin A\left( \dfrac{12\left( 1-{{\sin }^{2}}A \right)-3}{\cos A\left( 4{{\cos }^{2}}A-3 \right)} \right) \\

& M=\sin A\left( \dfrac{9-12{{\sin }^{2}}A}{\cos A\left( 4{{\cos }^{2}}A-3 \right)} \right) \\

\end{align}$

Taking ‘3’ common from numerator, we get;

$\begin{align}

& M=3\sin A\left( \dfrac{3-4{{\sin }^{2}}A}{\cos A\left( 4{{\cos }^{2}}A-3 \right)} \right) \\

& or \\

& M=3\sin A\left( \dfrac{3\sin A-4{{\sin }^{3}}A}{4{{\cos }^{3}}A-3\cos A} \right) \\

\end{align}$

AS, we know trigonometric identities for converting $3\theta \ to\ \theta $ with sin and cos terms as follows;

$\begin{align}

& \sin 3A=3\sin A-4{{\sin }^{3}}A \\

& \cos 3A=4{{\cos }^{3}}A-3\cos A \\

\end{align}$

Hence, above relation of M can be re-written as;

$M=3\dfrac{\sin 3A}{\cos 3A}=3\tan 3A$

Hence, the value of tan A + tan (60 + A) – tan (60 – A) is 3 tan 3A.

So, option A is the correct answer.

Note: One can get confused with the formulae of sin 3A and cos 3A as mentioned in the solution.

Another approach for this question would be that we can use trigonometry identities;

$\begin{align}

& \tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} \\

& and \\

& \tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B} \\

\end{align}$

to simplify tan (60 + A) and tan (60 – A). Then simplify the results found to get the answer. Applying trigonometric identities always makes solutions flexible and shorter which is the key point of the solution.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE