Question

(CoseccA-sinA)(secA-cosA)=1/tanA+cotA.........prove L.H.S=R.H.S

Answer 1

Answer:

hi......

Step-by-step explanation:

ye Kya hai........

Answer 2

Answer:

$lhs= (\frac{1}{ \sin \: a} - \sin \: a)( \frac{1}{ \cos \: a } - \cos \: a ) \\ \\ = \frac{1 - { \sin }^{2} a}{ \sin \: a} \times \frac{1 - {cos}^{2}a }{cos \: a} \\ \\ = \frac{ {cos}^{2}a }{sin \: a} \times \frac{ {sin}^{2} a}{cos \: a} \\ \\ = sin\: a \: cos \: a \\ \\ \\ \\ rhs = \frac{1}{ \frac{ {tan}^{2} a + 1}{tan \: a} } \\ \\ = \frac{tan \: a}{ {sec}^{2}a } \\ \\ = \frac{sin \: a}{cos \: a} \times {cos}^{2} a \\ \\ = sin\: a \: cos \: a \\ \\ lhs = rhs$