| If \[\sin \theta =\frac{{{a}^{2}}-1}{{{a}^{2}}+1},\]then the value of\[\sec \theta +\tan \theta \]will be |
A) \[\frac{a}{\sqrt{2}}\]
B) \[\frac{a}{{{a}^{2}}+1}\]
C) \[\sqrt{2}\,\,a\]
D) \[a\]
Correct Answer: D
Solution :
| \[\sin \theta =\frac{{{a}^{2}}-1}{{{a}^{2}}+1}\] |
 |
| In \[\Delta ABC,\] |
| \[BC=\sqrt{{{(AC)}^{2}}-{{(AB)}^{2}}}\] |
| \[=\sqrt{{{({{a}^{2}}+1)}^{2}}-{{({{a}^{2}}-1)}^{2}}}\] |
| \[=\sqrt{{{a}^{4}}+1+2{{a}^{2}}-{{a}^{4}}-1+2{{a}^{2}}}\] |
| \[=\sqrt{4{{a}^{2}}}=2a\] |
| \[\therefore \]\[\sec \theta +tan\theta =\frac{{{a}^{2}}+1}{2a}+\frac{{{a}^{2}}-1}{2a}\] |
| \[=\frac{{{a}^{2}}+1+{{a}^{2}}-1}{2a}=\frac{2{{a}^{2}}}{2a}=a\] |
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