How do you solve: tanθ + secθ = 1 ?

4 Answers
P dilip_k
May 23, 2018

#tanθ + secθ = 1....[1] #

Now
We know #sec^2theta-tan^2theta=1#
#=>sectheta-tantheta=1/(sectheta+tantheta)=1/1#

#=>sectheta-tantheta=1.....[2]#

Adding [1] and [2] we get

#2sectheta=2#

#=>costheta=1#

#=>theta=2npi" where "n inZZ"#

Sean
May 23, 2018

#theta=+-kpi, (3pi)/2+2kpi, -pi/2-2kpi#

Explanation:

.

#tantheta+sectheta=1#

#sintheta/costheta+1/costheta=1#

#(sintheta+1)/costheta=1#

#sintheta+1=costheta#

#(sintheta+1)^2=cos^2theta#

#sin^2theta+2sintheta+1=1-sin^2theta#

#2sin^2theta+2sintheta=0#

#2(sin^2theta+sintheta)=0#

#sin^2theta+sintheta=0#

#sintheta(sintheta+1)=0#

#sintheta=0, :. theta=+-kpi#

#sintheta+1=0, :. sintheta=-1, :. theta=(3pi)/2+2kpi, -pi/2-2kpi#

Tazwar Sikder
May 23, 2018

#frac(pi)(2) pm 2 pi n, pi pm 2 pi n#; #n in ZZ#

Explanation:

We have: #tan(theta) + sec(theta) = 1#

#Rightarrow frac(sin(theta))(cos(theta)) + frac(1)(cos(theta)) = 1#

#Rightarrow frac(sin(theta) + 1)(cos(theta)) = 1#

#Rightarrow sin(theta) + 1 = cos(theta)#

#Rightarrow sin(theta) - cos(theta) = - 1#

Then, let's consider #sin(theta) - cos(theta) = R cos(theta + alpha)#.

We can expand the right-hand side using the compound angle identity for #cos(theta)#:

#Rightarrow sin(theta) - cos(theta) = R (cos(theta) cos(alpha) - sin(theta) sin(alpha))#

#Rightarrow sin(theta) - cos(theta) = R cos(theta) cos(alpha) - R sin(theta) sin(alpha)#

#Rightarrow sin(theta) - cos(theta) = R cos(alpha) cos(theta) - R sin(alpha) sin(theta)#

If we compare the coefficients of #sin(theta)# and #cos(theta)#, we get:

#Rightarrow 1 = - R sin(alpha) Rightarrow R sin(alpha) = - 1 " " " " " "# #(i)#

and

#Rightarrow - 1 = R cos(alpha) Rightarrow R cos(alpha) = - 1 " " " " " "# #(ii)#

If we divide #(i)# by #(ii)#, we get:

#Rightarrow frac(R sin(alpha))(R cos(alpha)) = frac(- 1)(- 1)#

#Rightarrow tan(alpha) = 1#

#Rightarrow alpha = frac(pi)(4)#

If we sum the squares of #(i)# and #(ii)#, we get:

#Rightarrow R^(2) sin^(2)(alpha) + R^(2) cos^(2)(alpha) = 1 + 1#

#Rightarrow R^(2) (cos^(2)(alpha) + sin^(2)(alpha)) = 2#

#Rightarrow R^(2) = 2#

#Rightarrow R = sqrt(2)#

So, we can write #sin(theta) - cos(theta)# in the form #sqrt(2) cos(theta + frac(pi)(4))#, i.e. in our original problem we get #sqrt(2) cos(theta + frac(pi)(4)) = - 1#:

#Rightarrow cos(theta + frac(pi)(4)) = - frac(1)(sqrt(2)) = - frac(sqrt(2))(2)#

Now, let the reference angle be #cos(theta) = frac(sqrt(2))(2) Rightarrow theta = frac(pi)(4)#.

But the value of #cos(theta + frac(pi)(4))# is negative, so #theta# must be located in either the second or third quadrant:

#Rightarrow theta + frac(pi)(4) = pi - frac(pi)(4), pi + frac(pi)(4)#

#Rightarrow theta + frac(pi)(4) = frac(3 pi)(4), frac(5 pi)(4)#

#Rightarrow theta = frac(pi)(2), pi#

As you have not specified a domain that #theta# must be located within, we can give the general solutions to the equation as:

#therefore theta = frac(pi)(2) pm 2 pi n, pi pm 2 pi n#; #n in ZZ#

Nghi N
May 23, 2018

#t = kpi#
#t = (3pi)/2 + 2kpi#

Explanation:

#sin t/(cos t) + 1/cos t = 1 #
#(sin t + 1)/(cos t) = 1#
#sin t + 1 = cos t#
#sin t - cos t = - 1#
Use trig identity: #sin t - cos t = - sqrt2cos (t + pi/4)#
In this case:
#sin t - cos t = - sqrt2cos (t + pi/4) = - 1#
#cos (t + pi/4) = 1/sqrt2 = sqrt2/2#
Trig table and unit circle give 2 solutions for t:
#(t + pi/4) = +- pi/4#
a. #t + pi/4 = pi/4#
t = 0, #t = pi#, and #t = 2pi#
General answer --> #t = kpi#
b. t + pi/4 = - pi/4
#t = - pi/2 + 2kpi#, or #t = (3pi)/2 + 2kpi# (co-terminal)

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