Question

The value of Sec(2nπ-θ)
a)Sec(nθ)
b)Cos(nθ)
c)Sec(θ)
d)-Sec(θ)
Which one is correct ​

Answer 1

Answer:

a

Step-by-step explanation:

0

Given U

n

=sin(nθ).sec

n

θ=cos(nθ).sec

n

θ

V

n

−V

n−1

+U

n−1

tanθ

V

n

=cos(nθ).sec

n

θ

V

n

=cos((n−1)θ+θ).sec

n−1

θ.secθ

V

n

=sec

n−1

θ[cos(n−1)θ.cosθ−sin(n−1)θ.sinθ].secθ

V

n

=sec

n−1

θ[cos(n−1)θ.cosθ.secθ−sin(n−1)θ.sinθ.secθ]

V

n

=sec

n−1

θ[cos(n−1)θ−sin(n−1)θ.

cosθ

sinθ

]

V

n

=sec

n−1

θ[cos(n−1)θ]−sec

n−1

θ[sin(n−1)θ.tanθ]

V

n

=cos(n−1)θ.sec

n−1

θ−sin(n−1)θ.sec

n−1

θ.tanθ

V

n

=V

n−1

−[sin(n−1)θ.sec

n−1

θ].tanθ

V

n

=V

n−1

−[U

n−1

]tanθ

V

n

−V

n−1

+U

n−1

.tanθ=0

Answer 2

Answer:

OPTION A Will be the answer.