Question

tan A + cot A = sec A . cosec A

Answer 1

$\frac{ \sin( \alpha ) }{ \cos( \alpha ) } + \frac{ \cos( \alpha ) }{ \sin( \alpha ) } \\ \frac{ \sin {}^{2} \alpha + \cos {}^{2} \alpha }{ \sin( \alpha ) \cos( \alpha ) } \\ \frac{1}{ \sin( \alpha ) \cos( \alpha ) } \\ \\ \\ rhs \: side \: \\ \frac{1}{ \cos( \alpha ) } . \frac{1}{ \sin( \alpha ) } \\ lhs = rhs \\ hence \: proved$
..........take A = alpha

Answer 2

First of all tsinA= sinA /cosA
Ans cot A = coscosecA
So substituting them in LHS we get
SinA/cosA + cosA/sinA
Then adding them we get
Sin square A+cos square A upon cosA into sinA
Now as sin sq. + cos sq. = 1
We get 1/sinAcosA
1/sinA = cosecA and
1/cos A = SecA
Substituting them we get
SecA . cosecA