If #tan^2theta = 1 - a^2#, then #sectheta + tan^3thetacsctheta = (2 - a^2)^n#. What is the value of #n#?

Question

Delilah Aguilar · Accepted Answer

#3/2# and #theta in Q_1#.

Let #t = tan theta#. Then. #t^2=1-a^2>=0 to |a|<=1# Now, #sec theta+tan^3theta csc theta# #=+-sqrt(1+t^2)+t^3(+-sqrt(I+1/t^2))# #=+-sqrt(1+t^2)(1+-t^2)# #=+-(1+t^2)^(3/2)#, for + sign in the second factor ( #csc theta > 0#) #=+-(2-a^2)^(3/2 )#, #-# sign is taken when #sec theta < 0# #=(2-a^2)^n to sec theta > 0 and# #n=3/2# and csc theta >0 to theta in Q_1#. I am sorry that I have involved #+-#. But, this is Mathematics. I have obtained an enforcing result #theta in Q_1#.

Emery Adkins · Answer

#1.5#

#a^2 = 1 - tan^2theta#

Substitute:

#sectheta + tan^3thetacsctheta = (2 - (1 - tan^2theta))^n#

Apply the following identities:

#•secbeta = 1/cosbeta# #•tan beta = sinbeta/cosbeta# #•cscbeta = 1/sinbeta#

#1/costheta + sin^3theta/cos^3theta xx 1/sintheta = (2 - (1 - tan^2theta))^n#

#1/costheta + sin^2theta/cos^3theta = (1 + tan^2theta)^n#

Put the left hand side on a common denominator and apply the identity #1 + tan^2beta = sec^2beta#.

#(cos^2theta + sin^2theta)/cos^3theta = (sec^2theta)^n#

Use the identity #sin^2beta + cos^2beta = 1#.

#1/cos^3theta = (sec^2theta)^n#

#sec^3theta = (sec^2theta)^n#

We can call #sectheta = x#.

#x^3 = (x^2)^n#

#x^3 = x^(2n)#

#3 = 2n#

#n = 1.5#

Hopefully this helps!

Samantha Abate · Answer

#n = 3/2#

#1/costheta+sin^3theta/cos^3theta 1/sintheta=1/costheta(1+tan^2theta)=(2-a^2)^n#

then

#costheta=(2-a^2)/(2-a^2)^n=(2-a^2)^(-(n-1))#

squaring

#cos^2theta=(2-a^2)^(-2(n-1))#

but from #tan^2theta=(1-a^2)->cos^2theta=(2-a^2)^(-1)#

arriving at

#(2-a^2)^(-2(n-1))=(2-a^2)^(-1)# so

#2(n-1)=1->n=3/2#

Carter Anderson · Answer

Given #tan^2theta=1-a^2# #1+tan^2theta=1+1-a^2# #sec^2theta=2-a^2# Now
#sectheta+tan^3thetacsctheta=(2-a^2)^n# #=>sectheta+tan^3thetaxx1/sintheta=(2-a^2)^n# #=>sectheta(1+tan^3thetaxxcostheta/sintheta)=(2-a^2)^n# #=>sectheta(1+tan^3thetaxx1/tantheta)=(sec^2theta)^n# #=>sectheta(1+tan^2theta)=(sec^2theta)^n# #=>secthetaxxsec^2theta=sec^(2n)theta# #=>sec^3theta=sec^(2n)theta# #=>2n=3# #:.n=3/2#