Trig Identities Help

**Maccabhoy** · May 5th 2009, 12:54 AM

Hi guys, im doing a Maths quiz, and have gone through the basic Trig Identity stuff, but I am stuck on 2 questions....Anyone give me an idea what to do?

Which of the following is equal to (1 − cos2y)(1 + cot2y)

A) cos(2 y) B) 1 C) sec^2y D) sin^2y

and

Which of the following is equal to

1/ (sin x X cos x) − (cos x /sin x)

A) − 1 / (sin(2 x))

B) cos^2x

C) 1

D tan x

E) 1/ sin(2 x)

F) None of the above

**Maccabhoy** · May 5th 2009, 03:12 AM

Okay, I got the answer of 1 for the first question..

Stuck on the second though..

Any ideas?

**Soroban** · May 5th 2009, 09:30 AM

Hello, Maccabhoy!

Here's the second problem . . .

Which of the following is equal to: . $\frac{1}{\sin x\cos x} - \frac{\cos x}{\sin x}$

. . $(A)\;\frac{-1}{\sin2x} \quad(B)\;\cos^2\!x \qquad (C)\;1\qquad (D)\;\tan x \qquad <br /> (E)\;\frac{1}{\sin2x} \qquad (F)\text{ None of these}$

$\text{We have: }\;\frac{1}{\sin x\cos x} - \frac{\cos x}{\sin x}\cdot{\color{blue}\frac{\cos x}{\cos x}} \;\;=\;\;\frac{\overbrace{1 - \cos^2\!x}^{\text{This is }\sin^2\!x}}{\sin x\cos x}$

. . . . . . $= \;\frac{\sin^2\!x}{\sin x\cos x} \;=\;\frac{\sin x}{\cos x} \;=\;\tan x$ . . . answer (D)

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