Question

Please simplify the following trigonometric identity.

Answer 1

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:

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

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

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

We know that $\frac{1}{\frac{a}{b} }$ = $\frac{b}{a}$, using the same property:

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

and we are done!

Answer 2

Answer:

cos a

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1  -sin (a)

Step-by-step explanation:

We know that sec a = 1/ cos (a)  and tan a = sin (a) / cos (a)

1

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1/ cos (a)  -sin (a) / cos (a)

Multiply the top and bottom by cos (a)

1* cos a

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( 1/ cos (a)  -sin (a) / cos (a)) * cos a

cos a

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1  -sin (a)