Question

Prove that $$\dfrac{{\cos ({{90}^0} + \theta )\sec ({{270}^0} + \theta )\sin ({{180}^0} + \theta )}}{{\cos ( - \theta )\cos (270 - \theta )\tan (180 + \theta )}} = - \cos ec\theta $$

Answer 1

Hint:
The question is in complex form, we need to apply formulas of trigonometry. There are four quadrants in trigonometry. First we need to find out the given angle in which quadrant. According to their quadrant we will apply proper values. For e.g. $$\cos ({90^0} + \theta )$$ means angle of cos in $${90^0} + $$ . As the first quadrant limits up to 90 degree, $$\cos ({90^0} + \theta )$$ will be in the second quadrant.

Complete step by step solution:
We know that $\cos ({{90}^0}+\theta) = - \sin \theta$, $\sec ({{270}^0} + \theta)=\cos ec\theta$, $\sin ({{180}^0} + \theta)= - \sin \theta$, $\cos ( - \theta )=\cos (\theta )$, $\cos (270 - \theta )=- \sin \theta$ and $\tan (180 + \theta )=\tan \theta$.
Let’s use these values in LHS.
LHS=
$$\eqalign{
  & \dfrac{{ - \sin \theta \times (\cos ec\theta ) \times ( - \sin \theta )}}{{\cos \theta \times ( - \sin \theta ) \times \tan \theta }} \cr
  & = \dfrac{{\dfrac{1}{{\sin \theta }} \times ( - \sin \theta )}}{{\cos \theta \times \dfrac{{\sin \theta }}{{\cos \theta }}}} = \dfrac{{ - 1}}{{\sin \theta }} \cr
  & = - \cos ec\theta \cr
  & \cr} $$
=RHS
Hence, we proved that LHS=RHS i.e. $$\dfrac{{\cos ({{90}^0} + \theta )\sec ({{270}^0} + \theta )\sin ({{180}^0} + \theta )}}{{\cos ( - \theta )\cos (270 - \theta )\tan (180 + \theta )}} = - \cos ec\theta $$

Note:
There are four quadrants in trigonometry, we identified the quadrant and put the correct values in the correct term. In first quadrant all angle are considered as $$\theta $$ or $${90^0} - \theta $$ . in second quadrant all angles are considered as $${90^0} + \theta $$ or $${180^0} - \theta $$ . in third quadrant all angles are considered as $${180^0} + \theta $$ or $$270 - \theta $$ . In the fourth quadrant all angles are considered as $${270^0} + \theta $$ .