Show that : (2 + cos^2 X)(1+ tan^2 X)= 3 + 2tan^2 X?

Question

Show that : (2 + cos^2 X)(1+ tan^2 X)= 3 + 2tan^2 X

Answer 1

(2 + cos^2 x)(1+ tan^2 x)
= 2 + 2tan^2x + cos^2x + cos^2x*tan^2x
= 2 + 2tan^2x + cos^2x (1 + tan^2x)
= 2 + 2tan^2x + 1/sec^2x * sec^2x
= 2 + 2tan^2x + 1
= 3 + 2tan^2x

Answer 2

Substitute sinx / cos x for tan x to obtain: -

(2 + cos²x)(1 + sin²x/cos²x)
= 2 + 2 sin²x/cos² x + (cos²x + sin²x)
= 2 + 2 tan²x + 1
= 3 + 2 tan²x as required.

Answer 3

multiply out
2+ 2tan^2 (x) + cos^2 (x) + sin^2(x)tan^2(x) = 3 + 2tan^2 (x)

Remember tan^2 (x) = sin^2 (x) / cos^2 (x)

Insert for tan^2 (x)
2+ 2tan^2 (x) + cos^2 (x) + sin^2(x)(sin^2 (x) / cos^2 (x)) = 3 + 2tan^2 (x) = 3 + 2tan^2

2+ 2tan^2 (x) + cos^2 (x) + sin^2(x) = 3 + 2tan^2

Remember sin^2 (x) + cos^2 (x) = 1

2+ 2tan^2 (x) + 1 = 3 + 2tan^2

QED

Answer 4

The cost of tan is free in a hot country. But in a cold country, the cost of tan is 3 x the number of times in the tanning salon.

Answer 5

LHS=
(2 + cos^2 X)(1+ tan^2 X)
=2+(cosX)^2+2(tanX)^2
+(cosX)^2*(tanX)
{multiplying out}
=2+(cosX)^2+2(tanX)^2
+(cosX)^2*(sinX)^2/(cosX)^2
{tanX=sinX/cosX}
=2+[(cosX)^2+(sinX)^2]
+2(tanX)^2
=2+1+2(tanX)^2{pythagoras}
=3+2(tanX)^2
=RHS

therefore,

(2 + cos^2 X)(1+ tan^2 X)
= 3 + 2tan^2 X

i hope that this helps

Answer 6

I hope I'm not doing someone's homework! 2(x+3)=x+7 2x+6=x+7 (multiply 2 by the equation in brackets) 2x+6-x=7 (move the x's to one side) 2x-x=7-6 (move the numbers to the other) x=1 (do the equation on each side)

Answer 7

(2+cos^2x)(1+tan^2x) (( 1+tan^2x=1/cos^2x))
=(2+cos^2x)(1/cos^2x)
=2/cos^2x+1
=2(1+tan^2x)+1
=2+2tan^2x+1=3+2tan^2x

Answer 8

(2+cos^2X)(1+tan^2X)
=2+cos^2X+2tan^2X+cos^2X*tan^2X
but
tan^2X=sin^2X/cos^2X
thus
=2+cos^2X+2tan^2X+sin^2X
and we know that
cos^2X+sin^2X=1
thus the final equation is
2+1+2tan^2X=3+2tan^2X

Answer 9

open university?

Answer 10

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