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# Evaluate $\dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }}$ is equal toA. 0B. 1C. 2D. 3

Last updated date: 04th Aug 2024
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Answer
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Hint: Here, we will first rewrite the tangent functions of the given expression in terms of sine and cosine function. We will use various trigonometric formulas and trigonometric ratios to simplify the given trigonometric function. Trigonometric Ratios of a Particular angle are the ratios of the sides of a right angled triangle with respect to any of its acute angle.

Formula Used:
We will use the following formula:
1. Trigonometric Ratio: $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
2. Trigonometric Identity: $\sin A\cos B - \cos A\sin B = \sin \left( {A - B} \right)$
3. Trigonometric Identity: $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
4. Trigonometric Co-Ratio: $\cot \theta = \dfrac{1}{{\tan \theta }}$

Complete Step by Step Solution:
The given Trigonometric function is $\dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }}$.
We know that Trigonometric Ratio: $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
Now, we will rewrite tangent in terms of sine and cosine, so we get
$\Rightarrow \dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }} = \dfrac{{\dfrac{{\sin 80^\circ }}{{\cos 80^\circ }} - \dfrac{{\sin 10^\circ }}{{\cos 10^\circ }}}}{{\dfrac{{\sin 70^\circ }}{{\cos 70^\circ }}}}$
By cross multiplying the terms in the numerator, so we get
$\Rightarrow \dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }} = \dfrac{{\dfrac{{\sin 80^\circ \cos 10^\circ }}{{\cos 10^\circ \cos 80^\circ }} - \dfrac{{\sin 10^\circ \cos 80^\circ }}{{\cos 10^\circ \cos 80^\circ }}}}{{\dfrac{{\sin 70^\circ }}{{\cos 70^\circ }}}}$
Now, by rewriting the equation, we get
$\Rightarrow \dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }} = \dfrac{{\dfrac{{\sin 80^\circ \cos 10^\circ - \sin 10^\circ \cos 80^\circ }}{{\cos 10^\circ \cos 80^\circ }}}}{{\dfrac{{\sin 70^\circ }}{{\cos 70^\circ }}}}$
By using the Trigonometric Identity $\sin A\cos B - \cos A\sin B = \sin \left( {A - B} \right)$, we get
$\Rightarrow \dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }} = \dfrac{{\dfrac{{\sin \left( {80^\circ - 10^\circ } \right)}}{{\cos 10^\circ \cos 80^\circ }}}}{{\dfrac{{\sin 70^\circ }}{{\cos 70^\circ }}}}$
$\Rightarrow \dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }} = \dfrac{{\dfrac{{\sin 70^\circ }}{{\cos 10^\circ \cos 80^\circ }}}}{{\dfrac{{\sin 70^\circ }}{{\cos 70^\circ }}}}$
Now, by rewriting the equation, we get
$\Rightarrow \dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }} = \dfrac{{\sin 70^\circ \cos 70^\circ }}{{\cos 10^\circ \cos 80^\circ \sin 70^\circ }}$
Cancelling out the common terms, we get
$\Rightarrow \dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }} = \dfrac{{\cos 70^\circ }}{{\cos 10^\circ \cos 80^\circ }}$
Now, by rewriting the equation, we get
$\Rightarrow \dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }} = \dfrac{{\cos \left( {80^\circ - 10^\circ } \right)}}{{\cos 10^\circ \cos 80^\circ }}$
Rewriting the equation $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$, we get
$\Rightarrow \dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }} = \dfrac{{\cos 80^\circ \cos 10^\circ + \sin 10^\circ \sin 80^\circ }}{{\cos 10^\circ \cos 80^\circ }}$
Now, by segregating the terms, we get
$\Rightarrow \dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }} = \dfrac{{\cos 80^\circ \cos 10^\circ }}{{\cos 10^\circ \cos 80^\circ }} + \dfrac{{\sin 10^\circ \sin 80^\circ }}{{\cos 10^\circ \cos 80^\circ }}$
We know that Trigonometric Ratio: $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
Now, by cancelling out the common terms and by using the Trigonometric Identity, we get
$\Rightarrow \dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }} = 1 + \tan 10^\circ \tan 80^\circ$
Rewriting the equation, we get
$\Rightarrow \dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }} = 1 + \tan 10^\circ \tan \left( {90^\circ - 10^\circ } \right)$
Now, by using the Trigonometric Ratio $\tan \left( {90^\circ - x} \right) = \cot x$, we get
$\Rightarrow \dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }} = 1 + \tan 10^\circ \cot 10^\circ$
We know that Trigonometric Co-Ratio: $\cot \theta = \dfrac{1}{{\tan \theta }}$
By using the Trigonometric Ratio, we get
$\Rightarrow \dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }} = 1 + \tan 10^\circ \cdot \dfrac{1}{{\tan 10^\circ }}$
Now, by cancelling out the common terms, we get
$\Rightarrow \dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }} = 1 + 1$
Simplifying the equation, we get
$\Rightarrow \dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }} = 2$
Therefore, the value of $\dfrac{{\tan 80^\circ - \tan 10^\circ }}{{\tan 70^\circ }}$ is equal to 2.

Thus, option (C) is the correct answer.

Note:
We know that Trigonometric Equation is defined as an equation involving the trigonometric ratios. Trigonometric identity is an equation which is always true for all the variables. We should know that we have many trigonometric identities which are related to all the other trigonometric equations. Trigonometric Ratios are used to find the relationships between the sides of a right angle triangle. Trigonometric Ratio and Trigonometric Inverse Ratio are always inverses to each other.