# 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

Last updated date: 26th Mar 2023

•

Total views: 306.3k

•

Views today: 5.83k

Answer

Verified

306.3k+ 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.

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