Question

Tan A + cot A = sec A cosec A

Answer 1

cot + tan 
=cos/sin + sin/cos 
= (cos ^2 + sin^2) / sincos 
=1/sincos 
= 1/sin X 1/cos 
=cosec X sec 

=cosec sec

Answer 2

Hence proved

Step-by-step explanation:

Given:

Tan A + Cot A = Sec A Cosec A

we know that,

$Tan=\frac{Sin}{Cos} \\\\Cot=\frac{Cos}{Sin} \\\\Sec=\frac{1}{Cos} \\\\Cosec=\frac{1}{Sin}$

So,

$\frac{SinA}{CosA} +\frac{CosA}{SinA}= \frac{1}{CosA}\times\frac{1}{SinA}$

Taking LCM of both side

$\frac{Sin^2A+Cos^2A}{SinACosA}=\frac{1}{SinACosA}$

we know that,

$Sin^2A+Cos^2A = 1$

So,

$\frac{1}{SinACosA}=\frac{1}{SinACosA}$

L.H.S = R.H.S

Hence proved

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