Question

Prove that: $\frac{\mathrm{sec\theta }+\mathrm{tan\theta }-1}{\mathrm{tan\theta }-\mathrm{sec\theta }+1}=\frac{\mathrm{cos\theta }}{1-\mathrm{sin\theta }}$
Or, Evaluate: $\frac{\mathrm{sec\theta }\mathrm{cosec}(90°-\mathrm{\theta })-\mathrm{tan\theta }\cot (90°-\mathrm{\theta })+{\sin }^{2}55°+{\sin }^{2}35°}{\tan 10°\tan 20°\tan 60°\tan 70°\tan 80°}$

Answer 1

$\mathrm{LHS}=\left(\frac{\mathrm{sec\theta }+\mathrm{tan\theta }-1}{\mathrm{tan\theta }-\mathrm{sec\theta }+1}\right)$
$=\frac{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)-\left({\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }\right)}{\left(\mathrm{tan\theta }-\mathrm{sec\theta }+1\right)}$ [Since, 1 = sec²θ - tan²θ]
$=\frac{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)\left[1-\left(\mathrm{sec\theta }-\mathrm{tan\theta }\right)\right]}{\left(\mathrm{tan\theta }-\mathrm{sec\theta }+1\right)}$
$=\frac{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)\left(\mathrm{tan\theta }-\mathrm{sec\theta }+1\right)}{\left(\mathrm{tan\theta }-\mathrm{sec\theta }+1\right)}=\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)$
$$\begin{gathered} =\left(\frac{1}{\mathrm{cos\theta }}+\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}\right) \\ =\frac{\left(1+\mathrm{sin\theta }\right)}{\mathrm{cos\theta }}\times \frac{\left(1-\mathrm{sin\theta }\right)}{\left(1-\mathrm{sin\theta }\right)} \end{gathered}$$
$=\frac{\left(1-{\sin }^2\mathrm{\theta }\right)}{\mathrm{cos\theta }\left(1-\mathrm{sin\theta }\right)}=\frac{{\cos }^2\mathrm{\theta }}{\mathrm{cos\theta }\left(1-\mathrm{sin\theta }\right)}=\frac{\mathrm{cos\theta }}{\left(1-\mathrm{sin\theta }\right)}$
$\mathrm{RHS}=\frac{\mathrm{cos\theta }}{\left(1-\mathrm{sin\theta }\right)}$
Hence, LHS = RHS

OR

$\frac{\mathrm{sec\theta cosec}\left(90°-\mathrm{\theta }\right)-\mathrm{tan\theta cot}\left(90°-\mathrm{\theta }\right)+{\sin }^{2}55°+{\sin }^{2}35°}{\tan 10°\tan 20°\tan 60°\tan 70°\tan 80°}$

$=\frac{{\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }+{\sin }^2\left(90°-35°\right)+{\sin }^{2}35°}{\tan 10°\tan 20°\tan 60°\tan \left(90°-20°\right)\tan \left(90°-10°\right)}$

$=\frac{1+{\cos }^{2}35°+{\sin }^{2}35°}{\tan 10°\tan 20°\tan 60°\times \cot 20°\times \cot 10°}$ [Since, sec²θ - tan²θ = 1]

$=\frac{1+1}{\tan 10°\tan 20°\times \sqrt{3}\times {\displaystyle \frac{1}{\tan 20°}}\times {\displaystyle \frac{1}{\tan 10°}}}$ [Since, cos²θ + sin²θ = 1]
$=\frac{2}{\sqrt{3}}$