How do you solve: tanθ + secθ = 1 ?

4 Answers
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"

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

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

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)