Question

Prove that

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

 

Answer 1

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