Question

(cosecA-sinA)(secA-cosA)=1/tanA+cotA​

Answer 1

Answer:

$( \cosec a -\sin a )(\sec a-\cos a) = \frac{1}{ \tan a \: + \cot a}$

Explanation:

Given:

$( \cosec a -\sin a )(\sec a-\cos a) = \frac{1}{ \tan a \: + \cot a}$

To Prove:

$LHS=RHS$

$LHS$

$Substituting \: \: cosec \: a \: \: as \: \: \frac{1}{sin \: a} \: \: and \: \: sec \: a \: \: as \: \: \frac{1}{cos \: a}$

$( \frac{1}{sin \: a} - sin \: a)( \frac{1}{cos \: a} - cos \: a)$

Taking LCM, we get:

$( \frac{1 - {sin}^{2}a }{sin \: a} )( \frac{1 - {cos}^{2}a }{cos \: a} )$

$We \: know \: that \:1- {sin}^{2} a \:is \: {cos}^{2} a \: and \: 1- {cos}^{2} a \: is \: {sin}^{2} a$

$( \frac{{cos}^{2} a}{sin \: a})( \frac{ {sin}^{2}a }{cos \: a} )$

After performing calculation, we get:

$cos \: a \times sin \: a$

$RHS$

$Substituting \: \: tan \: a \: \: as \: \: \frac{sin \: a}{cos \: a} \: \: and \: \:cot \: a \: \: as \: \: \frac{cos \: a}{sin \: a}$

$\frac{1}{ \frac{sin \: a}{cos \: a} + \frac{cos \: a}{sin \: a} }$

Taking LCM, we get:

$\frac{1}{ \frac{sin \: a \times sin \: a \: + \: cos \: a \times cos \: a }{cos \: a \: sin \: a} }$

$\frac{1}{ \frac{ {sin}^{2} \: a + {cos}^{2} \: a}{cos \: a \: sin \: a} }$

We know that:

${sin}^{2} \: a \: + \: {cos}^{2} \: a \: = \: 1$

$\frac{1}{\frac{1}{cos \: a \: sin \: a} }$

$sin \: a \: cos \: a$

$\therefore \: LHS=RHS$

$Hence \: \: Proved$

Additional Information:

$\odot \: \: While \: doing \: proving \: questions \: ,$

$we \: can \: use \: trigonometric \: identities$

$and\:trigonometric\:ratios\:to\:prove.$

Trigonometric Identities

${sin}^{2} a + {cos}^{2} a = 1$

${sec}^{2} a - {tan}^{2} a = 1$

${cosec}^{2} a - {cot}^{2} a = 1$

Trigonometric Ratios

$sin \: a = \frac{1}{cosec \: a}$

$cos \: a = \frac{1}{sec \: a}$

$tan \: a = \frac{1}{cot \: a}$

$cot \: a = \frac{1}{tan \: a}$

$sec \: a = \frac{1}{cos \: a}$

$cosec \: a = \frac{1}{sin \: a}$

$\odot \: \: In \: some \: questions, \: the \: RHS \: is$

$already \: solved \: and \: we \: have$

$to \: just \: solve \: and \: prove \: LHS$

$to \: be \: equal \: to \: RHS.$

$\odot \: \: But \: in \: some \: questions \: like$

$this \: one, \: RHS \: is \: not \: solved$

$and \: we \: have \: to \: solve \: both$

$LHS \: and \: RHS \: to \: prove \: their$

$equality.$