| Directions: Each of these questions contains two statements: Assertion [A] and Reason [R]. Each of these questions also has four alternative choices, any one of which is the correct answer. You have to select one of the codes [a], [b], [c] and [d] given below. |
| Assertion If \[\sec \theta +\tan \theta =x\], then the value of \[\sin \theta =\frac{{{x}^{2}}-1}{{{x}^{2}}+1}\] |
| Reason \[x+\frac{1}{x}=2\,\tan \theta \] and \[x-\frac{1}{x}=2\,\sec \theta \]. |
A) A is true, R is true; R is a correct explanation for A.
B) A is true, R is true; R is not a correct explanation for A.
C) A is true; R is false.
D) A is false; R is true.
Correct Answer: C
Solution :
| We have, |
| \[\left( \sec \theta +\tan \theta \right)\left( \sec \theta -\tan \theta \right)=1\] |
| \[\Rightarrow \,x\left( \sec \theta -\tan \theta \right)=1\] |
| \[\Rightarrow \,\,\sec \theta -\tan \theta =\frac{1}{x}\] |
| Thus, we have |
| \[\sec \theta +\tan \theta =x\] |
| And \[\sec \theta -\tan \theta =\frac{1}{x}\] |
| Adding and subtracting these two equations, we get |
| \[2\sec \theta =x+\frac{1}{x}\] and \[2\tan \theta =x-\frac{1}{x}\] |
| \[\Rightarrow \sec \theta =\frac{1}{2}\left( x+\frac{1}{x} \right)\] and \[\tan \theta =\frac{1}{2}\left( x-\frac{1}{x} \right)\] |
| Now, \[\sin \theta =\frac{\tan \theta }{\sec \theta }\] |
| \[\Rightarrow \sin \theta =\frac{\frac{1}{2}\left( x-\frac{1}{x} \right)}{\frac{1}{2}\left( x+\frac{1}{x} \right)}=\frac{{{x}^{2}}-1}{{{x}^{2}}+1}\] |
| Assertion is true but Reason is false. |
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