Question

answer this question........ N plzzz explain it.

Answer 1

can I ask in which class you are

Answer 2

Given
$\frac{ \sin ^{4} x }{2} + \frac{ \cos ^{4} x }{3} = \frac{1}{5}$
Using sin²@+cos²@=1

Cos²@=1-sin²@

We get

$\frac{ \sin ^{4} x }{2} + \frac{(1 - \sin^{2}x )^{2} }{3} = \frac{1}{5}$
Let put sinx= a

$\frac{a ^{4} }{2} + \frac{(1 - a ^{2}) ^{2} }{3} = \frac{1}{5}$
$\frac{3a ^{4} + 2(1 - a ^{2}) ^{2} }{6} = \frac{1}{5}$
15a⁴+10(1-a²)²=6

Using (x-y) ²=x²+y²-2xy

15a⁴+10(1²+a⁴-2a²)=6

15a⁴+10+10a⁴-20a²-6=0

25a⁴-20a²+4=0

Using (x-y) ²= x²-2xy+y²

(5a²)²-2x2(5a²) +2²=0

(5a²-2) ²=0

5a²-2=0

a²=2/5

Replacing a² with sinx

Sin²x=2/5

1-cos²x=2/5

Cos²x=(2/5) - 1

Cos²x=3/5

Tan²x=sin²x/cos²x

=(2/5)/(3/5)

=2/3

Tan²x=2/3

So A=2

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