Question

cot=a/b prove that 2sec+1/cos+2=root of a^²+b^²/b

Answer 1

please mark on brain list Answer:

proof given below:

Explanation:

Taking conjugate of the denominator

#(1-cosTheta)/(1+cosTheta)=(1-cosTheta)/(1+cosTheta)*(1-cosTheta)/(1-cosTheta)#

#=(1-cosTheta)^2/(1^2-cos^2Theta)#

We have #sin^2Theta+cos^2Theta=1#
so #sin^2Theta=1-cos^2Theta#

#=(1-cosTheta)^2/sin^2Theta#

#=((1-cosTheta)/(sinTheta))^2#

Seperating the fractions

#=(1/sinTheta-cosTheta/sinTheta)^2#

#=(cscTheta-cotTheta)^2#

 

82



Shreya Naik

Sep 8, 2015

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Answer:

For solving this problem we need to know some trigonometric ratios and identities:
1)cosec#theta#=1/sin#theta#
2)cot#theta#=cos #theta#/sin#theta#
3)#sin^2##theta+cos^2##theta=1#

Explanation:

For an easy solution to all proof questions its convenient to go for 
L.H.S= R.H.S method:

1)in R.H.S (WE HAVE) #rArr# #(cosec##theta#- #cot##theta##)^2#

2) using 1 & 2 #rArr# #(1/sin##theta#-cos#theta#/#sin##theta##)^2#

3) now we get #rArr# #(1-cos##theta#/#sin##theta##)^2#

4) on splitting we get #rArr# (#1-cos##theta)^2# / (#sin##theta^2#)

5)now (#1-cos##theta)^2# can be written as :
(#1-cos##theta#) * # (1-cos##theta#)# and (#sin#theta^2#) can be written as :

(#1-cos##theta^2#) = (#1-cos##theta#) * # (1+cos##theta#)#

6) so,we get
(#1-cos##theta#) * (#1-cos##theta#) / (#1-cos##theta#) * (#1+cos##theta#)

7) Finally on cancelling we get , 
(#1-cos##theta#) / (#1+cos##theta#).......(so L.H.S= R.H.S)........... (Proved)
.......................................................................................................................
I hope this helps :)

 

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