Need help urently.

**lalitchugh** · June 12th 2006, 11:22 PM

Hi,

I need to solve below Qs. Can anyone help ASAP.

1.
Simplify: tan(180 – A) sec(90 + A) cosec (90 + A)
__________________________________
Sec (-A) cot (90 + A)

2. a) sec 120 b) sec 350 c) tan 40

Please mail me the solution on my mail id, lalitchugh@yahoo.com

Thanks in advance
Lalit Chugh

**Soroban** · June 13th 2006, 08:41 AM

Hello, lalitchugh!

I'll assume you're familiar with the Compound-Angle Identities . . .

$1)\;\text{Simplify: }\:\,\frac{\tan(180-A)\cdot\sec(90+A)\cdot\csc(90+A)}{\sec(-A)\cdot\cot(90+A)}$

I'll simplify each factor first . . .

$\tan(180 - A)\;=\;\frac{\tan180 - \tan A}{1 + \tan180\cdot\tan A} \;=$ $\;\frac{0 - \tan A}{1 + 0\cdot\tan A} \;= \;-\tan A$

$\sec(90 + A) \,=\,\frac{1}{\cos(90 + A)} \,=\,\frac{1}{\cos90\cos A - \sin90\sin A} \,= \,$ $\frac{1}{0\cdot\cos A - 1\cdot\sin A} \,= \,\frac{1}{\text{-}\sin A}\,=\,\text{-}\csc A$

$\csc(90+A) \,=\,\frac{1}{\sin(90+A)}\,=\,\frac{1}{\sin90\cos A + \cos90\sin A} \,=$ $\frac{1}{1\cdot\cos A + 0\cdot\sin A} \,=$ $\,\frac{1}{\cos A} \,= \,\sec A$

$\sec(-A) \,= \,\sec(A)$

$\cot(90 - A)\,=\,\frac{1 + \cos90\cos A}{\cot90 - \cot A} \,= \,\frac{1 + 0\cdot\cot A}{0 - \cot A}\,=$ $\,\frac{1}{\text{-}\cot A}\,=\,-\tan A$

The problem becomes: . $\frac{[-\tan A]\,[-\csc A]\,[\sec A]}{[\sec A]\,[-\tan A]} \;=\;-\csc A$

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