$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} =$

  • [IIT 1974]
  • A

    $0$

  • B

    $1$

  • C

    $2$

  • D

    $4$

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  • [JEE MAIN 2020]

Prove that $\cos ^{2} 2 x-\cos ^{2} 6 x=\sin 4 x \sin 8 x$

If $A + B + C = \pi ,$ then ${\tan ^2}\frac{A}{2} + {\tan ^2}\frac{B}{2} + $${\tan ^2}\frac{C}{2}$ is always

$1 + \cos \,{56^o} + \cos \,{58^o} - \cos {66^o} = $

  • [IIT 1964]

Prove that $\sin ^{2} 6 x-\sin ^{2} 4 x=\sin 2 x \sin 10 x$