If #p=cottheta+costheta# and #q=cottheta-costheta#, find the value of #p^2-q^2#?

Question

Abigail Armstrong · Accepted Answer

#p^2-q^2=4sqrt(pq)# As #p=cottheta+costheta# and #q=cottheta-costheta# #p^2-q^2# = #(p+q)(p-q)# = #2cotthetaxx2costheta# = #4cotthetacostheta# and #pq=cot^2theta-cos^2theta# = #cos^2theta/sin^2theta-cos^2theta# = #cos^2theta(csc^2theta-1)# = #cos^2thetacot^2theta# and #4costhetacottheta=4sqrt(pq)# Hence, #p^2-q^2=4sqrt(pq)#

Josiah Anderson · Answer

Given #p = cottheta+costheta# #q = cottheta-costheta# #pq=cot^2theta-cos^2theta# #=>pq=cos^2theta/sin^2theta-cos^2theta# #=>pq=cos^2theta(1/sin^2theta-1)#
#=>pq=cos^2theta(csc^2theta-1)# #=>pq=cos^2thetacot^2theta# #=>cotthetacostheta=sqrt(pq)# Now #p^2-q^2# #=(cottheta+costheta)^2- (cottheta-costheta)^2# #=4cotthetacostheta# #=4sqrt(pq)#

Victoria Adebayo · Answer

#p^2-q^2 = 4sqrt(pq)#

Square both p and q:

#p^2 = cot^2(theta)+ 2cot(theta)cos(theta) + cos^2(theta)" [1]"# #q^2 = cot^2(theta)- 2cot(theta)cos(theta) + cos^2(theta)" [2]"#

Subtract equation [2] from [1]:

#p^2-q^2 = 4cot(theta)cos(theta)" [3]"#

The right side of equation [3] is not in the selection list, therefore, we shall try to discover a connection by multiplying p and q:

#pq = (cot(theta)+cos(theta))(cos(theta)-cos(theta))" [4]"#

We know that the right of equation [4] is the difference of two squares:

#pq = cot^2(theta)-cos^2(theta)" [5]"#

Substitute #cos^2(theta)/sin^2(theta)# for #cot^2(theta)#:

#pq = cos^2(theta)/sin^2(theta)-cos^2(theta)" [6]"#

Make a common denominator:

#pq = cos^2(theta)/sin^2(theta)-(sin^2(theta)cos^2(theta))/sin^2(theta)" [7]"#

Remove #cos^2(theta)# as a common factor:

#pq = cos^2(theta) (1/sin^2(theta)-sin^2(theta)/sin^2(theta))" [8]"#

Combine over the common denominator:

#pq = cos^2(theta) ((1-sin^2(theta))/sin^2(theta))" [9]"#

Substitute #cos^2(theta)# for #1-sin^2(theta)#:

#pq = cos^2(theta) cos^2(theta)/sin^2(theta)" [10]"#

Substitute #cot^2(theta)# for #cos^2(theta)/sin^2(theta)#:

#pq = cot^2(theta)cos^2(theta)" [11]"#

Use the square operator on both sides:

#sqrt(pq) = cot(theta)cos(theta)" [12]"#

This allows us to substitute #sqrt(pq)# for #cot(theta)cos(theta)# into equation [3]:

#p^2-q^2 = 4sqrt(pq)" [3.1]"#

This is one of the selections in the list.

Emery Adkins · Answer

undefined $$ p^2 - q^2 = (cot\theta + cos\theta)^2 - (cot\theta - cos\theta)^2 $$
$$ = (cot^2\theta + 2cot\theta cos\theta + cos^2\theta) - (cot^2\theta - 2cot\theta cos\theta + cos^2\theta) $$
$$ = cot^2\theta + 2cot\theta cos\theta + cos^2\theta - cot^2\theta + 2cot\theta cos\theta - cos^2\theta $$
$$ = 4cot\theta cos\theta $$