Question

Prove that

\[\frac{\sin(180^\circ + x) \cos(90^\circ + x) \tan(270^\circ - x) \cot(360^\circ - x)}{\sin(360^\circ - x) \cos(360^\circ + x) cosec( - x) \sin(270^\circ + x)} = 1\]

 

Answer 1

 LHS = \[\frac{\sin \left( 180^\circ + x \right)\cos\left( 90^\circ + x \right) \tan \left( 270^\circ - x \right)\cot \left( 360^\circ - x \right)}{\sin \left( 360^\circ - x \right)\cos\left( 360^\circ + x \right)cosec\left( - x \right) \sin \left( 270^\circ + x \right)} \]
\[ = \frac{\sin \left( 90 \times 2^\circ + x \right)\cos\left( 90^\circ \times 1 + x \right) \tan\left( 90^\circ \times 3 - x \right) \cot\left( 90^\circ \times 4 - x \right)}{\sin\left( 90^\circ \times 4 - x \right)\cos\left( 90^\circ \times 4 + x \right) cosec \left( - x \right) \sin \left( 90^\circ \times 3 + x \right)}\]
\[ = \frac{- \sin x \left[ - \sin x \right] \cot x\left[ - \cot x \right]}{\left[ - \sin x \right] \cos x \left[ - cosec x \right]\left[ - \cos x \right]}\]
\[ = \frac{\sin^2 x \cot^2 x}{\sin x cosec x \cos x \cos x}\]
\[ = \frac{\sin^2 x \times \frac{\cos^2 x}{\sin^2 x}}{\sin x \times \frac{1}{\sin x} \times \cos^2 x}\]
\[ = \frac{\cos^2 x}{\cos^2 x}\]
\[ = 1\]
 = RHS
Hence proved .