A) 2
B) - 4
C) 3
D) 4
Correct Answer: D
Solution :
[d]: Given that \[\tan \theta -\cot \theta =a\] and \[\sin \theta +\cos \theta =b\] Now, \[{{({{b}^{2}}-1)}^{2}}({{a}^{2}}+4)\] \[={{\{{{(sin\theta +cos\theta )}^{2}}-1\}}^{2}}\{{{(tan\theta -cot\theta )}^{2}}+4\}\] \[={{[1+sin2\theta -1]}^{2}}[ta{{n}^{2}}\theta +co{{t}^{2}}\theta -2+4]\] \[={{\sin }^{2}}2\theta (cose{{c}^{2}}\theta +se{{c}^{2}}\theta )\] \[=4{{\sin }^{2}}\theta {{\cos }^{2}}\theta \left[ \frac{1}{{{\sin }^{2}}\theta }+\frac{1}{{{\cos }^{2}}\theta } \right]=4\]
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