Question

ooooo... brainy vasiooooo....

jara madat karoo koi....

pls help me sort this out,koi hai jo issea prove kar sake☝️☝️☝️☝️☝️☝️☝️☝️☝️☝️☝️☝️☝️☝️☝️​

Answer 1

||✪✪ QUESTION ✪✪||

Prove that sec⁴A(1-sin⁴A) - 2Tan²A = 1

|| ✰✰ ANSWER ✰✰ ||

Taking LHS,

→ sec⁴A(1-sin⁴A) - 2Tan²A

→ sec⁴A - sec⁴A*sin⁴A - 2Tan²A

Putting secA = (1/cosA) now,

→ sec⁴A - sin⁴A(1/cos⁴A)-2tan²A

Now, putting (SinA/cosA) = TanA

→ sec⁴A - tan⁴A - 2tan²A

→ [ (sec²A)² - (tan²A)²] - 2tan²A

Using (a² - b²) = (a+b)(a-b) Now,

→ (sec²A + tan²A)(sec²A - tan²A) - 2tan²A

Putting (sec²A - tan²A) = 1 now,

→ sec²A + tan²A - 2tan²A

→ sec²A - tan²A

→ 1 = RHS.

✪✪ Hence Proved ✪✪

Answer 2

Answer:

$\rm Question:-$

Prove following:-

• $\rm \sec^4 A (1 - \sin^4 A) - 2\tan^2 A = 1$

Identities used:-

$\bullet{\boxed{\rm{ sec \theta = \dfrac{1}{cos \theta}}}}$

$\bullet{\boxed{\rm {tan \theta = \dfrac{Sin\theta}{Cos\theta}}}}$

$\bullet{\boxed{\rm {\sec^2 \theta - \tan^2 \theta = 1 }}}$

$\rule{200}{1}$

$\rm L.H.S. = \sec^4 A (1 - \sin^4 A) - 2\tan^2 A$

$\rm \: \: \: \: = \sec^4 A - \sec^4 A \times \sin^4 A - 2 \tan^2 A$

$\rm \: \: \: \: [\because \sec^4 A = \dfrac{1}{\cos^4 A}]$

$\rm \: \: \: \: = \dfrac{1}{\cos^4 A} - \dfrac{1}{\cos^4 A } \times \sin^4 A - 2 \tan^2 A$

$\rm \: \: \: \: = (\dfrac{1}{\cos^4 A}) (1 - \sin^4 A) - 2 \tan^2 A$

$\rm \: \: \: \: = \dfrac{1 - \sin^4 A}{\cos^4 A} - 2 \tan^2 A$

$\rm \: \: \: \: = \dfrac{1}{\cos^4 A} - \dfrac{\sin^4 A}{\cos^4 A} - 2 \tan^2 A$

$\rm \: \: \: \: [\because tan^4 A=\dfrac{\sin^4 A}{\cos^4 A}]$

$\rm \: \: \: \: = (\sec^4 A - \tan^4 A) - 2 \tan^2 A$

$\rm \: \: \: \: [\because {a}^{2} - {b}^{2} = (a + b)(a - b)]$

$\rm \: \: \: \: = (\sec^2 A + \tan^2 A)(\sec^2 A - \tan^2 A) - 2 \tan^2 A$

$\rm \: \: \: \: = (\sec^2 A + \tan^2 A) (1) - 2 \tan^2 A$

$\rm \: \: \: \: [\because \sec^2 A - \tan^2 A = 1]$

$\rm \: \: \: \: = \sec^2 A + \tan^2 A - 2 \tan^2 A$

$\rm \: \: \: \: = \sec^2 A - \tan^2 A$

$\rm \: \: \: \: [\because \sec^2 A - \tan^2 A = 1]$

$\rm \: \: \: \: = 1$

$\rm \: \: \: \: = RHS$

$\therefore\rm LHS = RHS \: \: [Hence \: proved]$