(sin^2B-tan^2B)/(1-sec^2B)?

Question

please help me with this

Answer 1

I presume you want to simplify it, though you didn't say that :P
With any trig expression like this, you have two options:
1) Look for some cute trick. This is usually what you are shown when you get given the answer. However, these are often tricky to find.
2) Turn everything into sin and cos, and simplify. Then use the fact that (sin x)^2 + (cos x)^2 = 1. This *always* works!

So, turn tan into sin/cos, and sec into 1/cos:
((sin B)^2 - (sin B)^2/(cos B)^2) / (1 - 1/(cos B)^2)
Now, multiply top and bottom by (cos B)^2:
((sin B)^2 * (cos B)^2 - (sin B)^2) / ((cos B)^2 - 1).
Now, factorise the top:
(sin B)^2 ((cos B)^2 - 1) / ((cos B)^2 - 1)
And cancel:
(sin B)^2.

Answer 2

Let's see.... 1 + tan² B = sec² B, so 1 - sec² B should = -tan² B.

That makes your fraction (sin² B - tan² B) / -tan² B, and when you divide it out you have

-sin² B / tan² B + 1.

Now tan² B = sin² B / cos² B, so simplifying the fraction gives you

- cos² B + 1, which is equal to
sin² B.

Answer 3

You know that 1 + tan^2 b = sec^2 b, so 1 - sec^2 b = - tan^2 b.

Then (sin^2 b - tan^2 b) / (1 - sec^2 b)
= (sin^2 b - tan^2 b) / -tan^2 b
= (sin^2 b / -tan^2 b) - (tan^2 b / -tan^2 b)
= (sin^2 b / -tan^2 b) + 1
= (sin^2 b / -(sin^2 b/cos^2 b)) + 1
= -cos^2 b + 1
= 1- cos^2 b
= sin^2 b

Answer 4

sin²B - tan²B
---------------- =
1 - sec²B

sin²B - sin²B/cos²B)
-------------------------- =
(1 - 1/cos²B)


sin²B(1 - 1/cos²B)
----------------------- =
(1 - 1/cos²B)

sin²B