prove it

**saha.subham** · June 2nd 2010, 10:53 PM

sin theta(1 + tan theta) + cos theta(1 + cot theta) = sec theta + cosec theta

plzz complete it as i am new to this chapter

**simplependulum** · June 2nd 2010, 11:26 PM

Originally Posted by saha.subham

sin theta(1 + tan theta) + cos theta(1 + cot theta) = sec theta + cosec theta

plzz complete it as i am new to this chapter

It is just for easier input , i put $\theta = x$ so what we are going to prove is :

$\sin{x} ( 1 + \tan{x} ) + \cos{x} ( 1+ \cot{x} ) = \sec{x} + \csc{x}$

Consider

$\tan{x} = \frac{\sin{x}}{\cos{x}}$ and

$\cot{x} = \frac{\cos{x}}{\sin{x}}$

So

$1 + \tan{x} = 1 + \frac{\sin{x}}{\cos{x}} = \frac{ \sin{x} + \cos{x} }{ \cos{x}}$

$1 + \cot{x} = 1 + \frac{\cos{x}}{\sin{x}} = \frac{ \sin{x} + \cos{x} }{ \sin{x}}$

and we have

$\sin{x} ( 1 + \tan{x} ) + \cos{x} ( 1+ \cot{x} ) =$

$\sin{x}~\frac{ \sin{x} + \cos{x} }{ \cos{x}} + \cos{x} ~ \frac{ \sin{x} + \cos{x} }{ \sin{x}}$

Look at the common factor : $\sin{x} + \cos{x}$ get them out and we obtain :

$(\sin{x} + \cos{x}) \left(\frac{ \sin{x}}{\cos{x}} + \frac{ \cos{x}}{\sin{x}} \right)$

$= (\sin{x} + \cos{x}) ~ \frac{ \sin^2{x} + \cos^2{x} }{\sin{x}\cos{x}}$

I think you can finish it starting from here .

**nikhil** · June 3rd 2010, 12:22 AM

sinx(1+tanx)+cosx(1+cotx)
=sinx(1+tanx)+cosx[(1+tanx)/tanx]............................ [cotx=1/tanx]
=(1+tanx)[sinx+(cosx/tanx)]
=(1+tanx)[(sinx)^2+(cosx)^2]/sinx............................. [tanx=sinx/cosx]
=(1+tanx)/sinx
=((1/sinx)+(tanx/sinx))
=secx+cosecx

**Prove It** · June 3rd 2010, 12:28 AM

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