Question

Please solve question no. 99

Answer 1

$\underline{\underline{\mathfrak{\green{Answer:-}}}}$

(C) $\dfrac{1}{{x}^{2}}$

$\underline{\underline{\mathfrak{\green{Explanation:-}}}}$

Given:

secA + tanA = x

To Find:

$\mathsf{{sec}^{4}A-{tan}^{4}A-2\: secA \times tanA}$

Solution:

Given that,

$\mathsf{secA + tanA = x }$

As we know

$\boxed{\pink{{sec}^{2}A-{tan}^{2}A= 1}}$

Now,

$\mathsf{{sec}^{2}A-{tan}^{2}A = 1}$

$\mathsf{(secA+tanA)(secA-tanA)= 1}$

$\mathsf{x(secA-tanA)= 1}$

$\mathsf{secA-tanA= \dfrac{1}{x}}$

Let us find,

$\mathsf{{sec}^{4}A-{tan}^{4}A-2 \: secA \times tanA}$

_______________________

$\mathsf{={sec}^{4}A-{tan}^{4}A-2 \: secA \times tanA}$

$\mathsf{={({sec}^{2}A)}^{2}-{({tan}^{2}A)}^{2}-2 \: secA \times tanA}$

As we know

$\boxed{\pink{{a}^{2}-{b}^{2} = (a+b)(a-b)}}$

$\mathsf{=({sec}^{2}A+{tan}^{2}A)({sec}^{2}A-{tan}^{2}A)-2 \: secA \times tanA}$

$\mathsf{=({sec}^{2}A+{tan}^{2}A)(1)-2b\: secA \times tanA}$

$\mathsf{={sec}^{2}A+{tan}^{2}A-2 \: secA \times tanA}$

It is in the form-

$\boxed{\pink{{a}^{2}+{b}^{2}-2ab = {(a-b)}^{2}}}$

$\mathsf{={(secA-tanA)}^{2}}$

we got

$\boxed{\mathsf{secA-tanA= \dfrac{1}{x}}}$

By sub. the value

$\mathsf{={(\dfrac{1}{x})}^{2}}$

$\mathsf{= \dfrac{1}{{x}^{2}}}$

Hence,

$\mathsf{{sec}^{4}A-{tan}^{4}A-2 \: secA \times tanA = \dfrac{1}{{x}^{2}}}$

$\mathbb{\pink{NOTE}}$

Assume 'A' as 'α'

Answer 2

ANSWER:

given

$sec \alpha + tan \alpha = x$

REQUIRED TO FIND

${sec}^{4} \alpha - {tan}^{4} \alpha - 2sec \alpha \times tan \alpha$

METHOD

in the attachment

IDENTITIES USED:

$> {sec}^{2} \alpha - {tan}^{2} \alpha = 1 \\ \\ > {a}^{2} - {b}^{2} = (a + b)(a - b)$

$\\ > {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab$