Question

The general solution of the equation `"cot"  theta - "tan"  theta = "sec"  theta` is ____________ where `(n in I).`

Options

• `2 n pi + pi/6`

• `n pi + pi/6`

• `n  pi + (-1)^n pi/6`

• `2 n pi pm pi/6`

Answer 1

The general solution of the equation `"cot"  theta - "tan"  theta = "sec"  theta` is `underline(n  pi + (-1)^n pi/6)` where `(n in I).`

Explanation:

`"cot"  theta - "tan"  theta = "sec"  theta`

`=> ("cos" theta)/("sin" theta) - ("sin" theta)/("cos" theta) = 1/"cos" theta`

`=> ("cos"^2 - "sin"^2 theta)/("sin" theta. "cos" theta) = 1/"cos" theta`

`=> "cos"^2 theta - "sin"^2 theta = "sin" theta`

`=> "cos" 2 theta = "sin" theta`

`=> "sin" (pi/2 - 2 theta) = "sin" theta`

`=> pi/2 - 2 theta = theta => theta = pi/6`

General solution is `theta = n pi + (-1)^2 pi/6`