Question

Prove it.
Trignometry
Class 10

Answer 1

2(sin^6A+cos^6A) - 3(sin^4A+cos^4A) + 1

>> Convert 1 to (sin^2A + cos^2A)

= 2sin^6A + 2cos^6A - 3sin^4A - 3cos^4A
+ sin^2A + cos^2A

= 2sin^6A + 2cos^6A - 2sin^4A - 2cos^4A -sin^4A - cos^4A + sin^2A + cos^2A

= 2sin^6A - 2sin^4A + 2cos^6A - 2cos^4A
+ sin^2A - sin^4A + cos^2A - cos^4A

= -2sin^4A (1-sin^2A) - 2cos^4A (1-cos^2A)
+ sin^2A (1-sin^2A) + cos^2A (1-cos^2A)

= -2sin^4A cos^2A - 2cos^4A sin^2A + sin^2A cos^2A + cos^2A sin^2A

= -2sin^2A cos^2A (sin^2A+cos^2A)
+ 2sin^2Acos^2A

= -2sin^2A cos^2A + 2sin^2A cos^2A

=0

Answer 2

$\huge{\pink{\bold{\boxed{\ulcorner{\star\: Solution\: \star}\urcorner}}}}$

(A) Method

$2(sin^{6}\theta+cos^{6}\theta)-3(sin^{4}\theta+cos^{4}\theta)= -1$

$2((sin^{2}\theta)^{3}+(cos^{2}\theta)^{3})-3((sin^{2}\theta)^{2}+(cos^{2}\theta)^{2}))= -1$

Let
$a = sin^{2}\theta \\ b= cos^{2}\theta$

Let LHS

$2( a^{3} + b^{3} ) - 3( a^{2} + b^{2} )\\ 2[ (a+b)(a^{2} -ab+b^{3})]-3[(a+b)^{2}-2ab]\\2[ (a+b)((a+b)^{2}-2ab-ab)]-3[(a+b)^{2}-2ab]\\2[ (a+b)((a+b)^{2}-3ab)]-3[(a+b)^{2}-2ab]$

Put the value of

a & b in identity

$2[(sin^{2}\theta + cos^{2}\theta)((sin^{2}\theta+cos^{2}\theta)^{2}-3sin^{2}\theta.cos^{2}\theta)]-3[(sin^{2}\theta+cos^{2}\theta)^{2} -2sin^{2}\theta.cos\theta]$

$2[(1^{2})((1)^{2} - 3sin^{2}\theta.cos^{2}\theta)]-3[(1)^{2} -2sin^{2}\theta.cos\theta]\\ 2[1-3sin^{2}\theta.cos^{2}\theta]-3[ 1- 2sin^{2}\theta.cos^{2}]\\ 2- \cancel{6 sin^{2}\theta.cos^{2}\theta }- 3+\cancel{6sin^{2}\theta.cos^{2}\theta}\\ 2-3 \\ \boxed{\implies -1}$

------------------------------------------------------------------------------------------------------------------

(B) Method

$let \:1 \:= sin^{2}\theta+cos^{2}\theta$

$2(sin^{6}\theta+cos^{6}\theta)-3(sin^{4}\theta+cos^{4}\theta)+1 = 0$

$2((sin^{2}\theta)^{3}+(cos^{2}\theta)^{3})-3((sin^{2}\theta)^{2}+(cos^{2}\theta)^{2}))+sin^{2}\theta+cos^{2}\theta=1$

Let
$a = sin^{2}\theta \\ b= cos^{2}\theta$

Let LHS

$2( a^{3} + b^{3} ) - 3( a^{2} + b^{2} )\\ 2[ (a+b)(a^{2} -ab+b^{3})]-3[(a+b)^{2}-2ab]+[a+b]\\2[ (a+b)((a+b)^{2}-2ab-ab)]-3[(a+b)^{2}-2ab]+[a+b]\\2[ (a+b)((a+b)^{2}-3ab)]-3[(a+b)^{2}-2ab]+[a+b]$

Put the value of

a & b in identity

$2[(sin^{2}\theta + cos^{2}\theta)((sin^{2}\theta+cos^{2}\theta)^{2}-3sin^{2}\theta.cos^{2}\theta)]-3[(sin^{2}\theta+cos^{2}\theta)^{2} -2sin^{2}\theta.cos\theta]+sin^{2}\theta+cos^{2}\theta$

$2[(1^{2})((1)^{2} - 3sin^{2}\theta.cos^{2}\theta)]-3[(1)^{2} -2sin^{2}\theta.cos\theta]+sin^{2}\theta+cos^{2}\theta \\ 2[1-3sin^{2}\theta.cos^{2}\theta]-3[ 1- 2sin^{2}\theta.cos^{2}]+[sin^{2}\theta+cos^{2}\theta]\\ 2- \cancel{6 sin^{2}\theta.cos^{2}\theta }- 3+\cancel{6sin^{2}\theta.cos^{2}\theta}+sin^{2}\theta+cos^{2}\theta \\ 2-3 +1\\ 3-3 \\ \boxed{\implies 0}$

Now answer is

$\red{\boxed{\huge{LHS \:= \:RHS}}}$

Hence Proved