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TRIGONOMETRIC FUNCTIONS — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTRIGONOMETRIC FUNCTIONS1 Mark
Question
Prove that: $\frac { \cos ( \pi + x ) \cos ( - x ) } { \sin ( \pi - x ) \cos \left( \frac { \pi } { 2 } + x \right) } = \cot ^ { 2 } x$ .
✓

Answer

LHS = $\frac { \cos ( \pi + x ) \cdot \cos ( - x ) } { \sin ( \pi - x ) \cdot \cos \left( \frac { \pi } { 2 } + x \right) }$
= $\frac { - \cos x \cdot \cos x } { \sin x \cdot ( - \sin x ) }$
[$\because$ cos $( \pi + \theta )$ $= - cos\theta\  and\ cos (-\theta ) = cos\theta ]$
[$sin$ ($( \pi - \theta )$ $= sin$ $\theta$ and $cos$ $\left( \frac { \pi } { 2 } + \theta \right)$ $= - sin$ $\theta$]
= $\frac { \cos ^ { 2 } x } { \sin ^ { 2 } x }$$ = cot^2 x =$ RHS
Hence proved.

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