# Trigonometric Functions Short Questions (Question 11 – 14)

Question 11:
$\text{Prove the identity}\frac{{\mathrm{cos}}^{2}x}{1-\mathrm{sin}x}=1+\mathrm{sin}x$

Solution:

$\begin{array}{l}\text{LHS}\\ =\frac{{\mathrm{cos}}^{2}x}{1-\mathrm{sin}x}\\ =\frac{1-{\mathrm{sin}}^{2}x}{1-\mathrm{sin}x}←\overline{){\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1}\\ =\frac{\left(1+\mathrm{sin}x\right)\left(1-\mathrm{sin}x\right)}{1-\mathrm{sin}x}\\ =1+\mathrm{sin}x\\ =\text{RHS}\\ \therefore \text{Proven}\end{array}$

Question 12:
$\text{Prove the identity}{\mathrm{sin}}^{2}x-{\mathrm{cos}}^{2}x=\frac{{\mathrm{tan}}^{2}x-1}{{\mathrm{tan}}^{2}x+1}$

Solution:

$\begin{array}{l}\text{RHS}\\ =\frac{{\mathrm{tan}}^{2}x-1}{{\mathrm{tan}}^{2}x+1}\\ =\frac{\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}-1}{\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}+1}←\overline{)\mathrm{tan}x=\frac{\mathrm{sin}x}{\mathrm{cos}x}}\\ =\frac{\frac{{\mathrm{sin}}^{2}x-{\mathrm{cos}}^{2}x}{{\mathrm{cos}}^{2}x}}{\frac{{\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x}{{\mathrm{cos}}^{2}x}}\\ =\frac{{\mathrm{sin}}^{2}x-{\mathrm{cos}}^{2}x}{{\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x}\\ ={\mathrm{sin}}^{2}x-{\mathrm{cos}}^{2}x←\overline{){\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1}\\ =\text{LHS}\\ \therefore \text{Proven}\end{array}$

Question 13:
$\text{Prove the identity}{\mathrm{tan}}^{2}\theta -{\mathrm{sin}}^{2}\theta ={\mathrm{tan}}^{2}\theta {\mathrm{sin}}^{2}\theta$

Solution:

$\begin{array}{l}\text{LHS}\\ ={\mathrm{tan}}^{2}\theta -{\mathrm{sin}}^{2}\theta \\ =\frac{{\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta }-{\mathrm{sin}}^{2}\theta \\ =\frac{{\mathrm{sin}}^{2}\theta -{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }{{\mathrm{cos}}^{2}\theta }\\ =\frac{{\mathrm{sin}}^{2}\theta \left(1-{\mathrm{cos}}^{2}\theta \right)}{{\mathrm{cos}}^{2}\theta }\\ =\frac{{\mathrm{sin}}^{2}\theta {\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta }\\ =\left(\frac{{\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta }\right)\left({\mathrm{sin}}^{2}\theta \right)\\ ={\mathrm{tan}}^{2}\theta {\mathrm{sin}}^{2}\theta \\ =\text{RHS}\\ \therefore \text{Proven}\end{array}$

Question 14:
${\text{Prove the identity cosec}}^{2}\theta \text{}\left({\mathrm{sec}}^{2}\theta -{\mathrm{tan}}^{2}\theta \right)-1={\mathrm{cot}}^{2}\theta$

Solution:

$\begin{array}{l}\text{LHS}\\ ={\text{cosec}}^{2}\theta \text{}\left({\mathrm{sec}}^{2}\theta -{\mathrm{tan}}^{2}\theta \right)-1\\ ={\text{cosec}}^{2}\theta \text{}\left(1\right)-1←\overline{)\begin{array}{l}{\mathrm{tan}}^{2}\theta +1={\mathrm{sec}}^{2}\theta \\ {\mathrm{sec}}^{2}\theta -{\mathrm{tan}}^{2}\theta =1\end{array}}\\ ={\text{cosec}}^{2}\theta -1\\ ={\mathrm{cot}}^{2}\theta ←\overline{)\begin{array}{l}1+{\mathrm{cot}}^{2}\theta =\mathrm{cos}e{c}^{2}\theta \\ \mathrm{cos}e{c}^{2}\theta -1={\mathrm{cot}}^{2}\theta \end{array}}\\ =\text{RHS}\\ \therefore \text{Proven}\end{array}$