6.7.6 Trigonometric Functions Short Questions (Question 15 – 18)

Question 15:
$\text{Prove the identity}\frac{2}{\mathrm{cos}2A+1}=se{c}^{2}A$

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

Question 16:
$\text{Prove the identity}\frac{2\mathrm{tan}A}{2-se{c}^{2}A}=\mathrm{tan}2A$

Solution:
$\begin{array}{l}\text{LHS}\\ \text{=}\frac{2\mathrm{tan}A}{2-se{c}^{2}A}\\ =\frac{2\mathrm{tan}A}{2-\left({\mathrm{tan}}^{2}A+1\right)}←\overline{){\mathrm{tan}}^{2}A+1=se{c}^{2}A}\\ =\frac{2\mathrm{tan}A}{1-{\mathrm{tan}}^{2}A}\\ =\mathrm{tan}2A\\ =\text{RHS}\\ \therefore \text{Proven}\end{array}$

Question 17:
$\text{Prove the identity}\mathrm{tan}x+\mathrm{cot}x=2\mathrm{cos}ec\text{}2x$

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

Question 18:
$\text{Prove the identity}\frac{\mathrm{cos}x-\mathrm{sin}2x}{\mathrm{cos}2x+\mathrm{sin}x-1}=\frac{1}{\mathrm{tan}x}$

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