# If a straight line makes angles $\alpha ,\beta \text{,}\gamma $ with the coordinate axes then $\dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }+\dfrac{1}{\sec 2\beta }-2{{\sin }^{2}}\gamma $ is equal to.

A. -1

B. -2

C. 2

D. 0

Answer

Verified

361.2k+ views

Hint: First of all, use the formula ${{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1$. Convert ${{\cos }^{2}}\alpha $ in terms of $\cos 2\alpha $ by using the formula $\dfrac{1+\cos 2\alpha }{2}$. Then, convert $\cos 2\alpha $ in terms of ${{\tan }^{2}}\alpha $ by using the formula $\cos 2\alpha =\dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }$.

Complete step-by-step answer:

We have been given that a straight line makes angles $\alpha ,\beta ,\gamma $ with the coordinate axes.

Here, we have to use general formulas like $1+\cos 2\theta =2{{\cos }^{2}}\theta $,${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$,$\cos 2\theta =\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }$ and $\sec \theta =\dfrac{1}{\cos \theta }$.

Before proceeding with the question, we must know that the direction cosines of a line are given by $\cos \alpha ,\cos \beta ,\cos \gamma $and the sum of squares of the direction cosines of a line is equal to 1.

$\therefore {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1.....(i)$

We know that ${{\cos }^{2}}\alpha =\dfrac{1+\cos 2\alpha }{2}$ ,${{\cos }^{2}}\beta =\dfrac{1+\cos 2\beta }{2}$ . Therefore, we can substitute this value of ${{\cos }^{2}}\alpha $ and ${{\cos }^{2}}\beta $ in equation (i) as below,

$\Rightarrow \dfrac{1+\cos 2\alpha }{2}+\dfrac{1+\cos 2\beta }{2}+{{\cos }^{2}}\gamma =1.....(ii)$

We know that ${{\cos }^{2}}\gamma +{{\sin }^{2}}\gamma =1$ therefore, we can replace ${{\cos }^{2}}\gamma $as${{\cos }^{2}}\gamma =1-{{\sin }^{2}}\gamma $in equation (ii).

$\Rightarrow \dfrac{1+\cos 2\alpha }{2}+\dfrac{1+\cos 2\beta }{2}+1-{{\sin }^{2}}\gamma =1$

We can take LCM of denominators to perform the basic addition of fractions.

$\Rightarrow \dfrac{1+\cos 2\alpha +1+\cos 2\beta +2-2{{\sin }^{2}}\gamma }{2}=1$

Now, cross multiplying the above equation we get:

$\Rightarrow 1+\cos 2\alpha +1+\cos 2\beta +2-2{{\sin }^{2}}\gamma =2.....(iii)$

We know that $\cos 2\alpha =\dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }$ . Therefore, we can replace $\cos 2\alpha $ as $\dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }$ in equation (iii) and we will get,

$\Rightarrow 1+\dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }+1+\cos 2\beta +2-2{{\sin }^{2}}\gamma =2$

$\Rightarrow \dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }+4+\cos 2\beta -2{{\sin }^{2}}\gamma =2......(iv)$

We can replace $\cos 2\beta $ with $\dfrac{1}{\sec 2\beta }$ in the equation (iv) and we will get,

$\Rightarrow \dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\beta }+4+\dfrac{1}{\sec 2\beta }-2{{\sin }^{2}}\gamma =2$

$\Rightarrow \dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }+\dfrac{1}{\sec 2\beta }-2{{\sin }^{2}}\gamma =-2$

$\therefore \dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\beta }+\dfrac{1}{\sec 2\beta }-2{{\sin }^{2}}\gamma =-2$

Therefore, we have found the value of $\dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }+\dfrac{1}{\sec 2\beta }-2{{\sin }^{2}}\gamma $ as -2

Hence, the answer of the required question is option C.

Note: We have an alternate method for this question. We know that the maximum value of $\tan \alpha =\infty $ , $\sec \beta =\infty $ and ${{\sin }^{2}}\gamma =1$ . Therefore, just putting the maximum values of $\tan \alpha =\infty $,$\sec \beta =\infty $,${{\sin }^{2}}\gamma =1$ in the question, we get

$\begin{align}

& \dfrac{1}{\infty }+\dfrac{1}{\infty }-2 \\

& \\

\end{align}$

Therefore, we get the answer as -2, which is option C.

The whole question is concerned with angles and trigonometric identities. So, you must be able to recall all the formulas of trigonometric identities. If in question you are given any of the angles either $\alpha , \beta ,\gamma $, just check once by putting the values with their respective trigonometric ratios because in some cases we get the denominator as 0. In that case, if you have an option not defined then go for that option.

Complete step-by-step answer:

We have been given that a straight line makes angles $\alpha ,\beta ,\gamma $ with the coordinate axes.

Here, we have to use general formulas like $1+\cos 2\theta =2{{\cos }^{2}}\theta $,${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$,$\cos 2\theta =\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }$ and $\sec \theta =\dfrac{1}{\cos \theta }$.

Before proceeding with the question, we must know that the direction cosines of a line are given by $\cos \alpha ,\cos \beta ,\cos \gamma $and the sum of squares of the direction cosines of a line is equal to 1.

$\therefore {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1.....(i)$

We know that ${{\cos }^{2}}\alpha =\dfrac{1+\cos 2\alpha }{2}$ ,${{\cos }^{2}}\beta =\dfrac{1+\cos 2\beta }{2}$ . Therefore, we can substitute this value of ${{\cos }^{2}}\alpha $ and ${{\cos }^{2}}\beta $ in equation (i) as below,

$\Rightarrow \dfrac{1+\cos 2\alpha }{2}+\dfrac{1+\cos 2\beta }{2}+{{\cos }^{2}}\gamma =1.....(ii)$

We know that ${{\cos }^{2}}\gamma +{{\sin }^{2}}\gamma =1$ therefore, we can replace ${{\cos }^{2}}\gamma $as${{\cos }^{2}}\gamma =1-{{\sin }^{2}}\gamma $in equation (ii).

$\Rightarrow \dfrac{1+\cos 2\alpha }{2}+\dfrac{1+\cos 2\beta }{2}+1-{{\sin }^{2}}\gamma =1$

We can take LCM of denominators to perform the basic addition of fractions.

$\Rightarrow \dfrac{1+\cos 2\alpha +1+\cos 2\beta +2-2{{\sin }^{2}}\gamma }{2}=1$

Now, cross multiplying the above equation we get:

$\Rightarrow 1+\cos 2\alpha +1+\cos 2\beta +2-2{{\sin }^{2}}\gamma =2.....(iii)$

We know that $\cos 2\alpha =\dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }$ . Therefore, we can replace $\cos 2\alpha $ as $\dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }$ in equation (iii) and we will get,

$\Rightarrow 1+\dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }+1+\cos 2\beta +2-2{{\sin }^{2}}\gamma =2$

$\Rightarrow \dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }+4+\cos 2\beta -2{{\sin }^{2}}\gamma =2......(iv)$

We can replace $\cos 2\beta $ with $\dfrac{1}{\sec 2\beta }$ in the equation (iv) and we will get,

$\Rightarrow \dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\beta }+4+\dfrac{1}{\sec 2\beta }-2{{\sin }^{2}}\gamma =2$

$\Rightarrow \dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }+\dfrac{1}{\sec 2\beta }-2{{\sin }^{2}}\gamma =-2$

$\therefore \dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\beta }+\dfrac{1}{\sec 2\beta }-2{{\sin }^{2}}\gamma =-2$

Therefore, we have found the value of $\dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }+\dfrac{1}{\sec 2\beta }-2{{\sin }^{2}}\gamma $ as -2

Hence, the answer of the required question is option C.

Note: We have an alternate method for this question. We know that the maximum value of $\tan \alpha =\infty $ , $\sec \beta =\infty $ and ${{\sin }^{2}}\gamma =1$ . Therefore, just putting the maximum values of $\tan \alpha =\infty $,$\sec \beta =\infty $,${{\sin }^{2}}\gamma =1$ in the question, we get

$\begin{align}

& \dfrac{1}{\infty }+\dfrac{1}{\infty }-2 \\

& \\

\end{align}$

Therefore, we get the answer as -2, which is option C.

The whole question is concerned with angles and trigonometric identities. So, you must be able to recall all the formulas of trigonometric identities. If in question you are given any of the angles either $\alpha , \beta ,\gamma $, just check once by putting the values with their respective trigonometric ratios because in some cases we get the denominator as 0. In that case, if you have an option not defined then go for that option.

Last updated date: 29th Sep 2023

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