Question

2Q. $$\begin{gathered} \left(i\right)\sin \theta \cos \theta -\frac{\sin \theta \cos (90°-\theta )\cos \theta }{sec(90°-\theta )} \\ -\frac{\cos \theta \sin (90°-\theta )\sin \theta }{\cos ec(90°-\theta )}=0 \end{gathered}$$

Answer 1

Dear student
Q 2 i)
$$\begin{gathered} ConsiderLHS: \\ \mathrm{sin\theta cos\theta }-\frac{\mathrm{sin\theta cos}\left(90°-\mathrm{\theta }\right)\mathrm{cos\theta }}{\sec \left(90°-\mathrm{\theta }\right)}-\frac{\cos \theta \sin \left(90°-\mathrm{\theta }\right)\sin \theta }{co\sec \left(90°-\mathrm{\theta }\right)} \\ =\mathrm{sin\theta cos\theta }-\frac{\mathrm{sin\theta sin\theta cos\theta }}{\mathrm{cosec\theta }}-\frac{\cos \theta \cos \theta \sin \theta }{sec\theta }\left[\mathrm{as}\cos \left(90°-\mathrm{x}\right)=\mathrm{sinx},\sin \left(90°-\mathrm{x}\right)=\cos xandco\sec \left(90°-\mathrm{x}\right)=secx\right] \\ =\mathrm{sin\theta cos\theta }-\frac{\mathrm{sin\theta sin\theta cos\theta }}{{\displaystyle \frac{1}{\mathrm{sin\theta }}}}-\frac{\cos \theta \cos \theta \sin \theta }{{\displaystyle \frac{1}{\cos \theta }}}\left[\mathrm{cosecx}=\frac{1}{\mathrm{sinx}}andsecx=\frac{1}{\cos x}\right] \\ =\mathrm{sin\theta cos\theta }-{\sin }^2\mathrm{\theta sin\theta cos\theta }-{\cos }^3\theta \sin \theta \\ =\mathrm{sin\theta cos\theta }\left(1-{\sin }^2\mathrm{\theta }\right)-{\cos }^3\theta \sin \theta \\ ={\mathrm{sin\theta cos\theta cos}}^2\mathrm{\theta }-{\cos }^3\theta \sin \theta \left[\mathrm{as}{\cos }^2\mathrm{x}+{\sin }^2\mathrm{x}=1\right] \\ ={\mathrm{sin\theta cos}}^3\mathrm{\theta }-{\cos }^3\mathrm{\theta sin\theta } \\ =0 \\ =\mathrm{RHS} \end{gathered}$$
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