tan^2a-sin^2a=tan^2a sin ^2a, i tried showing this was an identity by using the (a+b) (a-b)and still cant fig.

Answer 1

take the LHS
u know that we can write tan^2 A = Sin^2 A/Cos^2 A
therefore taking Sin^2 A common and taking LCM
sin^2a/cos^2a - sin^2a (take LCM from this step)
sin^2 a(1-cos^2 a)/cos^2 a
u know sin over cos is tan
so
tan^2a(1-cos^2a)
and 1-cos^2a is sin^2a
therefore
tan^2a sin^2a

Answer 2

because tan^2 a is on the RHS do not disturb the 11st term on LHS

LHS = tan^2 a(1- sin^2 a/ tan^2 a)
= tan^2 a(1- cos^2 a) as sin a/tan a = cos a
= tan^2 a. sin^2 a (as sin^2 a + cos^2 a =1 so 1- cos^2 a = sin^2 a )

Answer 3

In this sort of problem, quite a sensible idea (as your attempt was) sometimes just isn't the right one, so you have to think of another one. Whenever I see tan and sin, I think of the fact that

tan a = (sin a)/(cos a),

or, to avoid writing fractions,

tan a = (sin a)*(sec a)

Use that here, and then both terms have (sin a)^2 as a factor, so you take out the common factor and then use the fact that

(sec a)^2 = 1 + (tan a)^2

Answer 4

The identity is:
tan^2a - sin^2a = tan^2a*sin^2a
Solving LHS = tan^2a - sin^2a first
sin^2a/cos^2a - sin^2a
sin^2a(1/cos^2a - 1)
sin^2a(1 -cos^2a)/cos^2a or
sin^2a/cos^2a(1 -cos^2a)
tan^2a*sin^2a = RHS
Hence proved

Answer 5

tan^2(A)-sin^2(A) = tan^2(A)*sin^2(A)

Divide both sides by tan^2(A):

1 - [sin^2(A)/tan^2(A)] = sin^2(A)

sin/tan = sin/[sin/cos} = cos

1 - cos^2(A) = sin^2(A),

1 = sin^2(A) + cos^2(A), a known identity

Answer 6

tan(A)^2 - sin(A)^2

tan(A)^2 = sec(A)^2 - 1 = (1/(cosA))^2 - 1 = (1 - cos(A)^2)/(cos(A)^2)

((1 - cosA^2)/(cosA^2)) - sinA^2

(1 - cosA^2 - (cosAsinA)^2)/(cosA^2)

sinA^2 = 1 - cosA^2

(1 - cosA^2 - (cosA^2 * (1 - cosA^2))/(cosA^2)

((1 - cosA^2)(1 - csoA^2))/(cosA^2)

((1 - cosA^2)^2)/(cosA^2)

((sinA^2)^2)/(cosA^2)

((sinA^2 * sinA^2)/(cosA^2)

(sinA^2 / cosA^2) * sinA^2

tanA^2 * sinA^2

so therefore

tanA^2 - sinA^2 = tanA^2 * sinA^2

Answer 7

(tanA)^2-(sinA)^2= sinA*(1/(cosA)^2-1)
as 1+(tanA)^2=1/(cosA)^2, thus
(sinA)^2*(1+(tanA)^2-1)= (tanA)^2*(sinA)^2