Question

A
Prove the following identities.
A + tan A cosec A - cot A
à
- tan
cosec A+ cot A
tan At co​

Answer 1

Question :

Prove that :-

$\sf\dfrac{SecA+TanA}{CosecA+CotA}$ = $\sf\dfrac{CosecA-CotA}{SecA-TanA}$

Solution :

We know that,

sec²A - Tan²A = 1

∵ (a + b)(a - b) = a² - b²

(SecA - TanA)(SecA + TanA) = 1

$\implies$ SecA + TanA = $\sf\dfrac{1}{SecA-TanA}$ _________(1)

Also,

Cosec²A - Cot²A = 1

∵ (a + b)(a - b) = a² - b²

(CosecA - CotA)(CosecA + CotA) = 1

$\implies$ CosecA + CotA = $\sf\dfrac{1}{CosecA-CotA}$ __________(2)

___________________

LHS :

Put the value of (1) and (2) in LHS,

$\implies$ $\sf\dfrac{SecA+TanA}{CosecA+CotA}$

$\implies$ $\sf\dfrac{\frac{1}{secA-TanA}}{\frac{1}{cosecA-CotA}}$

$\implies$ $\sf\dfrac{1}{SecA-TanA}$ × $\sf\dfrac{CosecA-CotA}{1}$

$\implies$ $\sf\dfrac{CosecA-CotA}{SecA-TanA}$

$\implies$ RHS

Hence proved!