Question

simplify
$\sf\frac {cos (90°-A).sec (180°-A).sin(180°-A)}{sin(90°+A).tan (90°+A).cosec (90°+A)}$​

Answer 1

Answer:

Solution:-

$\\\qquad\quad\displaystyle\sf {:}\leadsto \dfrac {cos (90°-A).sec (180°-A).sin(180°-A)}{sin (90°+A).tan (90°+A).cosec(90°+A)}$

$\\\qquad\quad\displaystyle\sf {:}\leadsto \dfrac {sinA.(-secA).sinA}{cosA.(-cotA).secA}$

$\\\qquad\quad\displaystyle\sf {:}\leadsto \dfrac {sin^2A.(-secA)}{cosA.\dfrac {cosA}{sinA}.(-secA)}$

$\\\qquad\quad\displaystyle\sf {:}\leadsto \dfrac {sin^2A.sinA}{cos^2A}$

$\\\qquad\quad\displaystyle\sf {:}\leadsto \dfrac{sin^3A}{cos^2A}\red {\bigstar}$

Answer 2

Answer:

$\huge\bold{Solution}$

__________________________

i)

$\cos(90 - a ) = \sec \: a$

ii)

$\sin(180 - a) = \sin \: a$

iii)

$\cot(360 - a) = -\cot a$

iv)

$\sec(180 + a) = - \sec\alpha$

v)

$\tan(90 + \alpha ) = - \cot\alpha$

vi)

$\sin( - \alpha ) = - \sin\alpha$

__________________________

Now,

LHS=cosec 90-a .sin 180-A.cot 360-A/sec 180+A.tan 90+A.sin(-A)

$=[secA \times sinA \times (-cotA)] \div [(-secA \times (-cotA) \times (-sinA)]$

After cancellation, we get

= 1

= RHS

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