Eliminate #x# in the following?

(a) #p=3tanx# and #q=3secx#
(b) #p=1+cosx# and #q=1-sinx#
(c) #p=2sinx# and #q=3secx#

3 Answers
Shwetank Mauria
Oct 8, 2017

(a) #q^2=9+p^2#
(b) #p^2+q^2-2p-2q+1=0#
(c) #p^2q^2+36=4q^2#

Explanation:

(a) As #p=3tanx# and #q=3secx#,

we have #tanx=p/3# and #secx=q/3#

and therefore as #sec^2x=1+tan^2x#,

#q^2/9=1+p^2/9# or #q^2=9+p^2#

(b) #p=1+cosx# and #q=1-sinx#,

we have #cosx=p-1# and #sinx=1-q#

and therefore as #sin^2x+cos^2x=1#,

#(1-q)^2+(p-1)^2=1# i..e. #p^2+q^2-2p-2q+1=0#

(c) As #p=2sinx# and #q=3secx#,

we have #sinx=p/2# and #secx=q/3# or #cosx=3/q#

and therefore as #sin^2x+cos^2x=1#,

#p^2/4+9/q^2=1# i.e. #p^2q^2+36=4q^2#

P dilip_k
Oct 8, 2017

(a)
Given

#3tanx=P =>tanx=P/3# and

#3secx=q=>secx=q/3#

#sec^2x-tan^2x=1#

#=>q^2/9-p^2/9=1#

(b)

#sin^2x+cos^2x=1#

#=>(1-q)^2+(p-1)^2=1#

(c)

#sin^2x+cos^2x=1#

#=>(p/2)^2+(3/q)^2=1#

#=>p^2/4+9/q^2=1#

Narad T.
Oct 10, 2017

The answers are #(a)p^2+9=q^2#, #(b)##(p-1)^2+(1-q)^2=1#, and #(c) #(3/q)^2+(p/2)^2=1##

Explanation:

We need

#cos^2x+sin^2x=1#

#1+tan^2x=sec^2x#

First part #(a)#

#p=3tanx#, #=>#, #tanx=p/3#

#q=3secx#, #=>#, #secx=q/3#

Therefore,

#1+tan^2x=sec^2x#

#1+(p/3)^2=(q/3)^2#

#p^2+9=q^2#

Second part #(b)#

#p=1+cosx#, #=>#, #cosx=p-1#

#q=1-sinx#, #=>#, #sinx=1-q#

Therefore,

#cos^2x+sin^2x=1#

#(p-1)^2+(1-q)^2=1#

Third part #(c)#

#p=2sinx#, #=>#, #sinx=p/2#

#q=3secx#, #=>#, #q=3/cosx#, #=>#, #cosx=3/q#

Therefore,

#cos^2x+sin^2x=1#

#(3/q)^2+(p/2)^2=1#

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