Question

Please simplify the following trigonometric identity.
\displaystyle\frac{1}{\sec\alpha-\tan\alpha}​

Answer 1

Answer:

Answer. Let a be the positive odd integer which when divided by 6 gives q as quotient and r as remainder. but here, a=6q+1 & a=6q+3 & a=6q+5 are odd.

Answer 2

Answer:

Examining the question:

We are given the expression:

$\frac{1}{Sec(\alpha)-Tan(\alpha)}$

We know from Basic trigonometry that:

$Sec(\alpha) = \frac{1}{Cos(\alpha)}$

$Tan(\alpha) = \frac{Sin(\alpha)}{Cos(\alpha)}$

Simplifying the expression:

Replacing these values in the given expression, we get:

Since the denominator of both the values in the denominator is the same:

Since the denominator of both the values in the denominator is the same:

$\frac{1}{\frac{1-Sin(\alpha)}{Cos(\alpha)} }$

$\frac{Cos(\alpha)}{1-Sin(\alpha)}$

and we are done!