Question

Prove that $\frac{\mathrm{sin\theta }-\mathrm{cos\theta }+1}{\mathrm{sin\theta }+\mathrm{cos\theta }-1}=\frac{1}{(\mathrm{sec\theta }-\mathrm{tan\theta })}.$
Or, evaluate: $\frac{\mathrm{sec\theta }\mathrm{cosec}(90°-\mathrm{\theta })-\tan \mathrm{\theta }\cot (90°-\mathrm{\theta })+{\sin }^{2}65°+{\sin }^{2}25°}{\tan 10°\tan 20°\tan 60°\tan 70°\tan 80°}$

Answer 1

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

$=\frac{{\displaystyle \frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}-1+{\displaystyle \frac{1}{\mathrm{cos\theta }}}}{{\displaystyle \frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}+1-{\displaystyle \frac{1}{\mathrm{cos\theta }}}}$ [On dividing numerator and denominator by cos θ]
$=\frac{\mathrm{tan\theta }-1+\mathrm{sec\theta }}{\mathrm{tan\theta }+1-\mathrm{sec\theta }}=\left(\frac{\mathrm{sec\theta }+\mathrm{tan\theta }-1}{\mathrm{tan\theta }-\mathrm{sec\theta }+1}\right)$
$=\frac{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)-\left({\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }\right)}{\left(\mathrm{tan\theta }-\mathrm{sec\theta }+1\right)}$ [Since 1 = sec²θ - tan²θ]
$=\frac{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)\left[1-\left(\mathrm{sec\theta }-\mathrm{tan\theta }\right)\right]}{\left(\mathrm{tan\theta }-\mathrm{sec\theta }+1\right)}$
$=\frac{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)\left(\mathrm{tan\theta }-\mathrm{sec\theta }+1\right)}{\left(\mathrm{tan\theta }-\mathrm{sec\theta }+1\right)}=\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)$

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

OR
$\frac{\mathrm{sec\theta cosec}\left(90°-\mathrm{\theta }\right)-\mathrm{tan\theta cot}\left(90°-\mathrm{\theta }\right)+{\sin }^{2}65°+{\sin }^{2}25°}{\tan 10°\tan 20°\tan 60°\tan 70°\tan 80°}$

$=\frac{{\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }+{\sin }^2\left(90°-25°\right)+{\sin }^{2}25°}{\tan 10°\tan 20°\tan 60°\tan \left(90°-20°\right)\tan \left(90°-10°\right)}$

$=\frac{1+{\cos }^{2}25°+{\sin }^{2}25°}{\tan 10°\tan 20°\tan 60°\times \cot 20°\times \cot 10°}$ [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]
$=\frac{2}{\sqrt{3}}$