The expression $\frac{{{{\tan }^2}20^\circ - {{\sin }^2}20^\circ }}{{{{\tan }^2}20^\circ \,\cdot\,{{\sin }^2}20^\circ }}$ simplifies to
The expression $\frac{{{{\tan }^2}20^\circ - {{\sin }^2}20^\circ }}{{{{\tan }^2}20^\circ \,\cdot\,{{\sin }^2}20^\circ }}$ simplifies to
- A
a rational which is not integral
- B
a surd
- C
a natural which is prime
- D
a natural which is not composite
Similar Questions
Prove that $\cos ^{2} 2 x-\cos ^{2} 6 x=\sin 4 x \sin 8 x$
Prove that $\cos ^{2} 2 x-\cos ^{2} 6 x=\sin 4 x \sin 8 x$
The value of $\tan 7\frac{1}{2}^\circ $ is equal to
The value of $\tan 7\frac{1}{2}^\circ $ is equal to
The exact value of $cos^273^o + cos^247^o + (cos73^o . cos47^o )$ is
The exact value of $cos^273^o + cos^247^o + (cos73^o . cos47^o )$ is
The value of $\sin \frac{\pi }{{14}}\sin \frac{{3\pi }}{{14}}\sin \frac{{5\pi }}{{14}}\sin \frac{{7\pi }}{{14}}\sin \frac{{9\pi }}{{14}}\sin \frac{{11\pi }}{{14}}\sin \frac{{13\pi }}{{14}}$ is equal to
The value of $\sin \frac{\pi }{{14}}\sin \frac{{3\pi }}{{14}}\sin \frac{{5\pi }}{{14}}\sin \frac{{7\pi }}{{14}}\sin \frac{{9\pi }}{{14}}\sin \frac{{11\pi }}{{14}}\sin \frac{{13\pi }}{{14}}$ is equal to
- [IIT 1991]
If $\frac{{2\sin \alpha }}{{\{ 1 + \cos \alpha + \sin \alpha \} }} = y,$ then $\frac{{\{ 1 - \cos \alpha + \sin \alpha \} }}{{1 + \sin \alpha }} = $
If $\frac{{2\sin \alpha }}{{\{ 1 + \cos \alpha + \sin \alpha \} }} = y,$ then $\frac{{\{ 1 - \cos \alpha + \sin \alpha \} }}{{1 + \sin \alpha }} = $