solve the equation tan theta + root 2 secant theta is equal to 1
Answers
Answer:
Answer:
\theta=2n\pi-\frac{\pi}{4}θ=2nπ−
4
π
Step-by-step explanation:
Given : Expression \sqrt 2\sec\theta+\tan \theta=1
2
secθ+tanθ=1
To find : Solve the expression ?
Solution :
Expression \sqrt 2\sec\theta+\tan \theta=1
2
secθ+tanθ=1
\sqrt 2\sec\theta=1-\tan \theta
2
secθ=1−tanθ
Squaring both side,
(\sqrt 2\sec\theta)^2=(1-\tan \theta)^2(
2
secθ)
2
=(1−tanθ)
2
2\sec^2\theta=(1-\tan \theta)^22sec
2
θ=(1−tanθ)
2
We know, \sec^2\theta=1+\tan^2\thetasec
2
θ=1+tan
2
θ
2+2\tan^2\theta=1+\tan^2\theta-2\tan\theta2+2tan
2
θ=1+tan
2
θ−2tanθ
\tan^2\theta+2\tan\theta+1=0tan
2
θ+2tanθ+1=0
(\tan\theta+1)^2=0(tanθ+1)
2
=0
Squaring both side,
\tan\theta+1=0tanθ+1=0
\tan\theta=-1tanθ=−1
\tan\theta=\tan (2n\pi-\frac{\pi}{4})tanθ=tan(2nπ−
4
π
)
\theta=2n\pi-\frac{\pi}{4}θ=2nπ−
4
π
Therefore, \theta=2n\pi-\frac{\pi}{4}θ=2nπ−
4
π