Question

Prove that: $\text{cos}\theta \text{sin}\theta -\frac{\text{cos}({90}^{\circ }-\theta )\text{cos}({90}^{\circ }-\theta )\text{cos}\theta }{\text{sec}({90}^{\circ }-\theta )}-\frac{\text{sin}({90}^{\circ }-\theta )\text{sin}({90}^{\circ }-\theta )\text{sin}\theta }{\text{cosec}({90}^{\circ }-\theta )}+\text{sec}({90}^{\circ }-\theta )=\frac{1}{\text{sin}\theta }$

Answer 1

LHS $=\text{cos}\theta \text{sin}\theta -\frac{\text{cos}({90}^{\circ }-\theta )\text{cos}({90}^{\circ }-\theta )\text{cos}\theta }{\text{sec}({90}^{\circ }-\theta )}-\frac{\text{sin}({90}^{\circ }-\theta )\text{sin}({90}^{\circ }-\theta )\text{sin}\theta }{\text{cosec}({90}^{\circ }-\theta )}+\text{sec}({90}^{\circ }-\theta )$

$=\text{cos}\theta \text{sin}\theta -\frac{\text{sin}\theta \text{sin}\theta \text{cos}\theta }{\text{cosec}\theta }-\frac{\text{cos}\theta \text{cos}\theta \text{sin}\theta }{\text{sec}\theta }+\text{cosec}\theta$

$=\text{cos}\theta \text{sin}\theta -\frac{{\text{sin}}^2\theta \text{cos}\theta }{\frac{1}{\text{sin}\theta }}-\frac{{\text{cos}}^2\theta \text{sin}\theta }{\frac{1}{\text{cos}\theta }}+\frac{1}{\text{sin}\theta }$

$=\text{cos}\theta \text{sin}\theta -{\text{sin}}^3\theta \text{cos}\theta -{\text{cos}}^3\theta \text{sin}\theta +\frac{1}{\text{sin}\theta }$

$=\cos \theta \sin \theta -\sin \theta \cos \theta ({\sin }^2\theta +{\cos }^2\theta )+\frac{1}{\sin \theta }$

$=\text{cos}\theta \text{sin}\theta -\text{cos}\theta \text{sin}\theta +\frac{1}{\text{sin}\theta }$

$=\frac{1}{\text{sin}\theta }=$ RHS Hence Proved

Identities used:

i) $\text{cosec}A=\frac{1}{\text{sin}A}$

ii) $\text{sin}({90}^{\circ }-A)=\text{cos}A$

iii) $\text{cos}({90}^{\circ }-A)=\text{sin}A$

iv) $\text{sec}({90}^{\circ }-A)=\text{cosec}A$

v) $\text{cosec}({90}^{\circ }-A)=\text{sec}A$

vi) ${\text{sin}}^{2}A+{\text{cos}}^{2}A=1$

vii) $\text{sec}A=\frac{1}{\text{cos}A}$