Question

((cosec A - cot A)2 +1)/
(sec A (cosec A - cot A))
= 2 cot A
​

Answer 1

Qᴜᴇsᴛɪᴏɴ :-

Prove :- {(cosecA - cotA)² +1}/(secA(cosecA - cotA) = 2cotA

Sᴏʟᴜᴛɪᴏɴ :-

Taking LHS,

→ {(cosecA - cotA)² +1}/(secA(cosecA - cotA)

Putting 1 = cosec²A - cot²A in Numerator,

→ {(cosecA - cotA)² +(cosec²A - cot²A)}/(secA(cosecA - cotA)

using a² - b² = (a + b)(a - b) Now,

→ {(cosecA - cotA)² +(cosecA - cotA)(cosecA + cotA)}/(secA(cosecA - cotA)

Taking (cosecA - cotA) common from Numerator,

→ (cosecA - cotA){(cosecA - cotA) + (cosecA + cotA)} / secA(cosecA - cotA)

→ {(cosecA - cotA) + (cosecA + cotA)} / secA

→ 2cosecA / secA

Putting cosecA = (1/sinA) & secA = (1/cosA) Now,

→ 2*(1/sinA) / (1/cosA)

→ (2/sinA) * (cosA/1)

→ (2cosA/sinA)

→ 2cotA = RHS (Proved).

Answer 2

QUESTION:

$\frac{( { \csc \: a - \cot \: a) }^{2} + 1}{ \sec \: a ( \csc \: a - \cot \: a \: ) } = 2 \cot(a)$

ANSWER:

We use the trigonometry identity here;

${ \csc( \alpha ) }^{2} - { \cot( \alpha ) }^{2} = 1 \\ \\ \csc( \alpha ) = \frac{1}{ \sin( \alpha ) } \\ \\ \sec( \alpha ) = \frac{1}{ \cos( \alpha ) } \\ \\ \\ \cot( \alpha ) = \frac{ \cos( \alpha ) }{ \sin( \alpha ) }$

other identity we used here :

${x}^{2} - {y}^{2} = (x + y)(x - y)$

NOW COME TO YOUR QUESTION :

TAKING LHS;

I LET A AS ALPHA

$\frac{( { \csc( \alpha ) - \cot( \alpha ) }^{2} + 1 }{ \sec( \alpha ) ( \csc( \alpha ) - \cot( \alpha ) }$

using the identity

$\frac{( { \csc( \alpha ) - \cot( \alpha ) }^{2} + ( { \csc( \alpha ) }^{2} - { \cot( \alpha ) }^{2} }{ \sec( \alpha ) ( \csc( \alpha ) - \cot( \alpha ) }$

$\frac{( { \csc( \alpha ) - \cot( \alpha ) }^{2} + ( \csc( \alpha ) - \cot( \alpha ) ( \csc( \alpha ) + \cot( \alpha ) }{ \sec( \alpha ) ( \csc( \alpha ) - \cot( \alpha ) }$

TAKING common here

$\frac{( \csc( \alpha ) - \cot( \alpha ) + ( \csc( \alpha ) + \cot( \alpha ) }{ \sec( \alpha ) }$

$\frac{2 \csc( \alpha ) }{ \sec( \alpha ) }$

$\frac{2 \times \frac{1}{ \sin( \alpha ) } }{ \frac{1}{ \cos( \alpha ) } } \\ 2 \times \frac{ \cos( \alpha ) }{ \sin( \alpha ) }$

$2 \cot( \alpha )$

LHS = RHS

HENCE PROVED