Question

(1+4x^2)cosA = 4x then show that cosecA + cotA = 1+2x/1-2x

Answer 1

Here is the answer May It Help.

Answer 2

Answer:

cosecA + cotA = $\frac{(1 + 2x)}{(1-2x)}$

Step-by-step explanation:

Given that

$(1+4x^{2})cosA=4x$

We can rearrange the above equation as

$cosA=\frac{4x}{(1+4x^{2})}$ ... (1)

As we know that

$cosA=\frac{Base}{Hyp}$ ... (2)

Comparing (1) and (2)

$Base=4x$

$Hyp=(1+4x^{2})$

Now by pythagorean theorem we can find that

$Perp=1-4x^{2}$

As we know

$cosecA=\frac{hyp}{perp}$

$cosecA=\frac{1+4x^{2}}{1-4x^{2}}$

$cotA=\frac{base}{perp}$

$cosecA=\frac{4x}{1-4x^{2}}$

Now

cosecA + cotA = $\frac{1+4x^{2}}{1-4x^{2}}+\frac{4x}{1-4x^{2}}$

cosecA + cotA = $\frac{1+4x^{2} + 4x}{1-4x^{2}}$

cosecA + cotA = $\frac{(1 + 2x)^{2}}{(1-2x)(1+2x)}$

cancelling the same terms in numerator and denominator

cosecA + cotA = $\frac{(1 + 2x)}{(1-2x)}$

Hence proved