#( 1+ cot(\theta )- csc(\theta ) )( 1+tan(\theta )+ sec(\theta ) )= 2 #?

3 Answers
sankarankalyanam · P dilip_k
Feb 15, 2018

#=> color(green)(2)#

Explanation:

#(1+cot theta - csc theta) (1 + tan theta + sec theta)#

#=> (1 + (cos theta / sin theta) - (1/sin theta) )(1 + (sin theta/ cos theta) + (1/cos theta))#

#=>((sin theta + cos theta - 1) (cos theta + sin theta + 1))/(sinthetacostheta)#

#=> ((sin theta + cos theta)^2 - 1)/(sinthetacostheta)#

#=> (cancel(sin^2 theta + cos^2 theta)^color(red)1+ 2 sin theta cos theta - 1)/(sinthetacostheta)#

#=>( cancel 1 + 2 sin theta cos theta - cancel1) /(sinthetacostheta)#

#=> color(green)(2 )#

P dilip_k
Feb 15, 2018

#LHS=(1+ cottheta-csc theta)(1+ tantheta+sectheta)#

#=(1+ cottheta-csc theta+ tantheta+ tantheta cottheta-tanthetacsctheta+sectheta+secthetacottheta-secthetacsctheta)#

#=(1+ cottheta-csc theta+ tantheta+ 1-sintheta/costheta*1/sintheta+sectheta+1/costheta*costheta/sintheta-secthetacsctheta)#

#=(2+ cottheta-csc theta+ tantheta-sectheta+sectheta+csctheta-secthetacsctheta)#

#=(2+ costheta/sintheta+ sintheta/costheta-secthetacsctheta)#

#=(2+ (cos^2theta+sin^2theta)/( sinthetacostheta)-secthetacsctheta)#

#=(2+ 1/( sinthetacostheta)-secthetacsctheta)#

#=(2+ cscthetasectheta-secthetacsctheta)#

#=2=RHS#

Sridhar V.
Feb 15, 2018

Please see the explanation.

Explanation:

Prove that

#color(red)(( 1+ cot(\theta )- csc(\theta ) )( 1+tan(\theta )+ sec(\theta ) )= 2#

First, we will list the trigonometric identities we will use:

#color(blue)(sin^2(\theta )+cos^2(\theta)-=1#

#color(blue)(sin(\theta )/cos(\theta)-=tan(\theta)#

#color(blue)(1/sin(\theta )-=csc(\theta)#

#color(blue)(1/cos(\theta )-=sec(\theta)#

#color(blue)(cot(\theta )-=1/tan\theta-=cos(\theta)/sin(\theta)#

We will consider the Left-Hand Side (LHS) first:

#([1+ cot(\theta )- csc(\theta ) )( 1+tan(\theta )+ sec(\theta ) ]#

#rArr (1+ 1/tan(\theta )- 1/sin(\theta ) )( 1+sin(\theta )/cos(\theta )+ 1/cos(\theta))#

#rArr (1/1+ cos(\theta )/sin(\theta )- 1/sin(\theta ) )( 1/1+sin(\theta )/cos(\theta )+ 1/cos(\theta))#

#rArr ((sin(\theta )+ cos(\theta )- 1 )/sin(\theta ) )((cos(\theta )+sin(\theta )+ 1)/cos(\theta))#

#rArr [((sin(\theta )+ cos(\theta )- 1 ) (cos(\theta )+sin(\theta )+ 1)]/(sin(\theta)*cos(\theta)]]#

#rArr (sin(\theta )*cos(\theta )+ sin^2(\theta )+cancel sin (\theta ))/(sin(\theta)*cos(\theta)]#+

#(cos^2(\theta )+sin(\theta)*cos(\theta )+cancel cos(\theta ))/(sin(\theta)*cos(\theta)]#+

#(-cancel cos(\theta )-cancel sin(\theta )- 1)/(sin(\theta)*cos(\theta)]#

#rArr (sin^2(\theta )+cos^2(\theta )+sin(\theta)*cos(\theta)+sin(\theta)*cos(\theta)-1)/(sin(\theta)*cos(\theta)#

#rArr (cancel 1+2*sin(\theta)*cos(\theta)- cancel 1)/(sin(\theta)*cos(\theta)#

#rArr (2*sin(\theta)*cos(\theta))/(sin(\theta)*cos(\theta)#

#rArr 2*cancel(sin(\theta)*cos(\theta))/cancel(sin(\theta)*cos(\theta)#

#rArr 2#

= Right-Hand Side (RHS)

Hence,

#color(blue)(( 1+ cot(\theta )- csc(\theta ) )( 1+tan(\theta )+ sec(\theta ) )= 2#

Related questions
Impact of this question
6464 views around the world
You can reuse this answer
Creative Commons License