Question

Prove that:
( 1 - tan^4 A)cos^4 A = 1 - 2 sin^2 A​

Answer 1

See proof below

Explanation:

Prove
`\left(\:1\:-\:tan^4A\right)cos^4A\:=\:1\:-\:2\:sin^2A`

`\left(1-\tan ^4\left(A\right)\right)\cos ^4\left(A\right)` can be expressed in sin, cos terms

Use the trigonometric identity
`\tan \left(x\right)=(\sin \left(x\right))/(\cos \left(x\right))`

`\left(1-\tan ^4\left(A\right)\right)\cos ^4\left(A\right) = \left(1-\left((\sin \left(A\right))/(\cos \left(A\right))\right)^4\right)\cos ^4\left(A\right)`

`\mathrm{Simplify}\:\left(1-\left((\sin \left(A\right))/(\cos \left(A\right))\right)^4\right)\cos ^4\left(A\right)`

`\mathrm{Apply\:exponent\:rule}:\quad \left((a)/(b)\right)^c=(a^c)/(b^c)`

=
`\left(1-(\sin ^4\left(A\right))/(\cos ^4\left(A\right))\right)\cos ^4\left(A\right)`

Multiplying the expression in parentheses by
`\cos ^4\left(A\right)` we get

`(\cos ^4\left(A\right)-\sin ^4\left(A\right))/(\cos ^4\left(A\right))\cos ^4\left(A\right)`

Cancel the common factor
`\cos ^4\left(A\right)`

This gives us

`\cos ^4\left(A\right)-\sin ^4\left(A\right)`

Now,

`\sin ^4\left(A\right)=\left(\sin ^2\left(A\right)\right)^2`

`\cos ^4\left(A\right)=\left(\cos ^2\left(A\right)\right)^2`

`\:\cos ^4\left(A\right)-\sin ^4\left(A\right)` =
`\left(\cos ^2\left(A\right)\right)^2-\left(\sin ^2\left(A\right)\right)^2`

`\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)`

`\left(\cos ^2\left(A\right)\right)^2-\left(\sin ^2\left(A\right)\right)^2=\left(\cos ^2\left(A\right)+\sin ^2\left(A\right)\right)\left(\cos ^2\left(A\right)-\sin ^2\left(A\right)\right)`

=
`\cos ^2\left(A\right)-\sin ^2\left(A\right)`
`\textrm{ since }\cos ^2\left(A\right)+\sin ^2\left(A\right)` = 1

Using the fact that
`\cos ^2\left(A\right)=1-\sin ^2\left(A\right)`

we get

`\cos ^2\left(A\right)-\sin ^2\left(A\right) = 1-\sin ^2\left(A\right)-\sin ^2\left(A\right)\\\\= 1-2\sin ^2\left(A\right)`

Proved

Answer 2

Here we go ~

`{ \qquad\qquad\huge\underline{{\sf Answer}}}`

`\qquad \sf  \dashrightarrow \: (1 - \tan {}^(4) (a) ) \cos {}^(4) (a)`

`\qquad \sf  \dashrightarrow \: (1 + \tan {}^(2) (a) )(1 - { \tan}^(2) (a)) \cos {}^(4)(a )`

[ a² - b² = (a + b)(a - b) ]

`\qquad \sf  \dashrightarrow \: ( \sec {}^(2) (a) )(1 - ( \sec{}^(2) (a) - 1) )\cos {}^(4) (a)`

[ sec² a = 1 + tan² a, so : tan² a = sec²a - 1 ]

`\qquad \sf  \dashrightarrow \: \bigg( \frac{1}{{}cos^(2) (a)} \bigg)(2 - \sec{}^(2) (a) ) \cos {}^(4) (a)`

`\qquad \sf  \dashrightarrow \: \bigg(2 - \frac{1}{ \cos {}^(2) (a) } \bigg) \cos {}^(2) (a)`

`\qquad \sf  \dashrightarrow \: \frac{2 \cos {}^(2) (a) - 1}{ \cos {}^(2) (a) } * \cos {}^(2) (a)`

`\qquad \sf  \dashrightarrow \: 2 \cos {}^(2) (a) - 1`

`\qquad \sf  \dashrightarrow \: 2(1 - \sin {}^(2) (a) ) - 1`

[ sin²a + cos² a = 1, hence sin²a = 1 - cos²a ]

`\qquad \sf  \dashrightarrow \: 2 - 2 \sin {}^(2) (a) - 1`

`\qquad \sf  \dashrightarrow \: 1 - 2 \sin {}^(2) (a)`