Question

Prove the following trigonometric identities.
(i) $\frac{1+\text{cos}\theta +\text{sin}\theta }{1+\text{cos}\theta -\text{sin}\theta }=\frac{1+\text{sin}\theta }{\text{cos}\theta }$

(ii) $\frac{\text{sin}\theta -\text{cos}\theta +1}{\text{sin}\theta +\text{cos}\theta -1}=\frac{1}{\text{sec}\theta -\text{tan}\theta }$

(iii) $\frac{\text{cos}\theta -\text{sin}\theta +1}{\text{cos}\theta +\text{sin}\theta -1}=\text{cosec}\theta +\text{cot}\theta$

(iv) $(\text{sin}\theta +\text{cos}\theta )(\text{tan}\theta +\text{cot}\theta )=\text{sec}\theta +\text{cosec}\theta$

Answer 1

(i) We have to prove the following identity-

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

Consider the LHS.

$\frac{1+\text{cos}\theta +\text{sin}\theta }{1+\text{cos}\theta -\text{sin}\theta }$
=$\left[\frac{1+\text{cos}\theta +\text{sin}\theta }{1+\text{cos}\theta -\text{sin}\theta }\right]\left[\frac{1+\text{cos}\theta +\text{sin}\theta }{1+\text{cos}\theta +\text{sin}\theta }\right]$

$=\frac{(1+\text{cos}\theta +\text{sin}\theta {)}^2}{(1+\text{cos}\theta {)}^2-{\text{sin}}^2\theta }$

$=\frac{(1+\text{cos}\theta {)}^2+{\text{sin}}^2\theta +2(1+\text{cos}\theta )\text{sin}\theta }{1+{\text{cos}}^2\theta +2\text{cos}\theta -{\text{sin}}^2\theta }$

$=\frac{1+{\text{cos}}^2\theta +2\text{cos}\theta +{\text{sin}}^2\theta +2\text{sin}\theta +2\text{sin}\theta \text{cos}\theta }{1+{\text{cos}}^2\theta +2\text{cos}\theta -{\text{sin}}^2\theta }$

$=\frac{2+2(\text{cos}\theta +\text{sin}\theta +\text{sin}\theta \text{cos}\theta )}{2{\text{cos}}^2\theta +2\text{cos}\theta }$ [$\because {\text{cos}}^2\theta +{\text{sin}}^2\theta =1$]

$=\frac{2(1+\text{cos}\theta +\text{sin}\theta +\text{sin}\theta \text{cos}\theta )}{2\text{cos}\theta (1+\text{cos}\theta )}$

$=\frac{2[1+\text{cos}\theta +\text{sin}\theta (1+\text{cos}\theta \left)\right]}{2\text{cos}\theta (1+\text{cos}\theta )}$

$=\frac{2(1+\text{cos}\theta )(1+\text{sin}\theta )]}{2\text{cos}\theta (1+\text{cos}\theta )}$

$=\frac{(1+\text{sin}\theta )}{\text{cos}\theta }=$ RHS

Hence proved.

(ii) We have to prove the following identity-

$\frac{\text{sin}\theta -\text{cos}\theta +1}{\text{sin}\theta +\text{cos}\theta -1}=\frac{1}{\text{sec}\theta -\text{tan}\theta }$

Consider the LHS.

$\frac{\text{sin}\theta -\text{cos}\theta +1}{\text{sin}\theta +\text{cos}\theta -1}$

$=\left[\frac{\text{sin}\theta -\text{cos}\theta +1}{\text{sin}\theta +\text{cos}\theta -1}\right]\left[\frac{\text{sin}\theta -\text{cos}\theta +1}{\text{sin}\theta +\text{cos}\theta +1}\right]$

$=\frac{(\text{sin}\theta +1{)}^2-{\text{cos}}^2\theta }{(\text{sin}\theta +\text{cos}\theta {)}^2-1}$

$=\frac{{\text{sin}}^2\theta +1+2\text{sin}\theta -{\text{cos}}^2\theta }{{\text{sin}}^2\theta +{\text{cos}}^2\theta +2\text{sin}\theta \text{cos}\theta -1}$

$=\frac{2{\text{sin}}^2\theta +2\text{sin}\theta }{2\text{sin}\theta \text{cos}\theta }$

$=\frac{2\text{sin}\theta (1+\text{sin}\theta )}{2\text{sin}\theta \text{cos}\theta }$

$=\frac{(1+\text{sin}\theta )}{\text{cos}\theta }$

$=\left[\frac{(1+\text{sin}\theta )}{\text{cos}\theta }\right]\left[\frac{(1-\text{sin}\theta )}{(1-\text{sin}\theta )}\right]$

$=\left[\frac{(1-{\text{sin}}^2\theta )}{\text{cos}\theta (1-\text{sin}\theta )}\right]$

$=\left[\frac{\left(\text{cos}\theta \right)}{(1-\text{sin}\theta )}\right]$

$=\left[\frac{1}{(\text{sec}\theta -\text{tan}\theta )}\right]=$RHS [Divide numerator and denominator by $\text{cos}\theta$]

Hence proved.

(iii) We have to prove the following identity-

$\frac{\text{cos}\theta -\text{sin}\theta +1}{\text{cos}\theta +\text{sin}\theta -1}=\text{cosec}\theta +\text{cot}\theta$

Consider the LHS.

$\frac{\text{cos}\theta -\text{sin}\theta +1}{\text{cos}\theta +\text{sin}\theta -1}$

$=\frac{\text{cos}\theta -\text{sin}\theta +1}{\text{cos}\theta +\text{sin}\theta -1}\times \frac{\text{cos}\theta +\text{sin}\theta +1}{\text{cos}\theta +\text{sin}\theta +1}$

$=\frac{(\text{cos}\theta +1{)}^2-{\text{sin}}^2\theta }{(\text{cos}\theta +\text{sin}\theta {)}^2-1^2}$

$=\frac{{\text{cos}}^2\theta +1+2\text{cos}\theta -{\text{sin}}^2\theta }{{\text{cos}}^2\theta +{\text{sin}}^2\theta +2\text{cos}\theta \text{sin}\theta -1}$

$=\frac{{\text{cos}}^2\theta +1+2\text{cos}\theta -(1-{\text{cos}}^2\theta )}{1+2\text{cos}\theta \text{sin}\theta -1}$

$=\frac{2{\text{cos}}^2\theta +2\text{cos}\theta }{2\text{cos}\theta \text{sin}\theta }$

$=\frac{2\text{cos}\theta (\text{cos}\theta +1)}{2\text{cos}\theta \text{sin}\theta }$

$=\frac{\text{cos}\theta +1}{\text{sin}\theta }$

$=\frac{\text{cos}\theta }{\text{sin}\theta }+\frac{1}{\text{sin}\theta }$

$=\text{cot}\theta +\text{cosec}\theta$= RHS

Hence proved.
(iv)
$(\text{sin}\theta +\text{cos}\theta )(\text{tan}\theta +\text{cot}\theta )=\text{sec}\theta +\text{cosec}\theta$

Consider the LHS.

$(\text{sin}\theta +\text{cos}\theta )(\text{tan}\theta +\text{cot}\theta )$

$=(\text{sin}\theta +\text{cos}\theta )(\frac{\text{sin}\theta }{\text{cos}\theta }+\frac{\text{cos}\theta }{\text{sin}\theta })$

$=(\text{sin}\theta +\text{cos}\theta )\left(\frac{{\text{sin}}^2\theta +{\text{cos}}^2\theta }{\text{sin}\theta \text{cos}\theta }\right)$ [$\because {\text{sin}}^2\theta +{\text{cos}}^2\theta =1$]

$=(\text{sin}\theta +\text{cos}\theta )\frac{1}{\text{sin}\theta \text{cos}\theta }$

$=\frac{(\text{sin}\theta +\text{cos}\theta )}{\left(\text{sin}\theta \text{cos}\theta \right)}$

$=\frac{1}{\text{cos}\theta }+\frac{1}{\text{sin}\theta }$

$=\text{sec}\theta +\text{cosec}\theta$
$=$ RHS
Hence proved.