Question

Which of the following option makes the statement below true?
{(1/ secx) + secx} /(cos^2x - 1 - tan²x)
(a)-cosecx.cotx (b)-secx.cotx (c)-cosecx.tanx (d) – secx.tanx​

Answer 1

Answer:

Answer in abobe attachment

Answer 2

Answer:

The simplification of the trigonometric function is $- cosecx . cotx$

Explanation:

Simply put, trigonometric functions—also referred to as circular functions—are the functions of a triangle's angle. This means that these trig functions provide the relationship between the angles and sides of a triangle.

Sine (sin), cosine (cos), tangent (tan), secant (sec), cosecant (cosec), and cotangent are the names of these six functions (cot).

There are different identities, that we have used to simplify the expressions.

The given expression is

$\frac{\frac{1}{secx} + secx}{(cosx)^{2} - 1 - (tanx)^{2} }$

We know, $\frac{1}{secx } = cosx$

And

$1 +(tanx)^{2}$ = $(secx)^{2}$

Now, Evaluating the expression, we have,

$\frac{\frac{1}{secx} + secx}{(cosx)^{2} - 1 - (tanx)^{2} } = \frac{cosx + secx}{(cosx)^{2} - (1 + (tanx)^{2}) }$

$\frac{cosx + secx}{(cosx)^{2} -(secx)^{2} } = \frac{cosx + secx}{(cosx - secx)(cosx + secx) } = \frac{1}{(cosx - secx)}$

Now, $\frac{1}{secx } = cosx$

So, $\frac{1}{cosx -\frac{1}{cosx} } = \frac{-cosx}{1 -(cosx)^{2} } = \frac{-cosx}{(sinx)^{2} }$

Now,$\frac{1}{cosecx } = sinx$

and $\frac{sinx}{cosx} = tanx$

So, We get,

$\ \frac{-cosx}{(sinx)^{2} } = - cosecx . cotx$

Hence, The value of the trigonometric function is $- cosecx . cotx$

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