A) \[\sqrt{{{x}^{2}}-1}\]
B) \[\sqrt{1-{{x}^{2}}}\]
C) \[{{x}^{2}}-1\]
D) \[1-{{x}^{2}}\]
Correct Answer: C
Solution :
(c): \[\frac{1}{\cos e{{c}^{2}}{{51}^{{}^\circ }}}+si{{n}^{2}}{{39}^{{}^\circ }}+ta{{n}^{2}}{{51}^{{}^\circ }}-\frac{1}{{{\sin }^{2}}{{51}^{{}^\circ }}{{\sec }^{2}}{{39}^{{}^\circ }}}\] \[=\sin {{25}^{{}^\circ }}si{{n}^{2}}{{39}^{{}^\circ }}ta{{n}^{2}}({{90}^{{}^\circ }}-{{39}^{{}^\circ }})-\frac{1}{{{\sin }^{2}}\left( {{90}^{{}^\circ }}-{{39}^{{}^\circ }} \right).{{\sec }^{2}}{{39}^{{}^\circ }}}\] \[=co{{s}^{2}}{{39}^{{}^\circ }}+{{\sin }^{2}}{{39}^{{}^\circ }}+co{{t}^{2}}{{39}^{{}^\circ }}-\frac{1}{{{\cos }^{2}}{{39}^{{}^\circ }}{{\sec }^{2}}{{39}^{{}^\circ }}}\] \[\left[ \therefore sin\left( {{90}^{{}^\circ }}-\theta \right)=cos\theta \text{ }tan\left( {{90}^{{}^\circ }}-\theta \right)=cot\theta \right]\] \[=1+co{{t}^{2}}{{39}^{{}^\circ }}-1=co{{t}^{2}}{{39}^{{}^\circ }}\] \[=cosec{{39}^{{}^\circ }}-1={{x}^{2}}-1\]
You need to login to perform this action.
You will be redirected in
3 sec
