Question

Prove that $(tan\theta +cot\theta {)}^2=se{c}^2\theta +cose{c}^2\theta$ [3 MARKS]

Answer 1

Formula: 2 Marks
Application: 1 Mark

Consider L.H.S

$(tan\theta +cot\theta {)}^2$

$(a+b{)}^2=a^2+b^2+2ab$

$=ta{n}^2\theta +co{t}^2\theta +2tan\theta cot\theta$

$=ta{n}^2\theta +co{t}^2\theta +2\left(1\right)$ $(\because cot\theta =\frac{1}{tan\theta })$

$=(se{c}^2\theta -1)+(cose{c}^2\theta -1)+2$ $(\because se{c}^2\theta -1=ta{n}^2\theta )$

$=se{c}^2\theta +cose{c}^2\theta -2+2$ $(\because cose{c}^2\theta -1=co{t}^2\theta )$

$=se{c}^2\theta +cose{c}^2\theta$