Simplify [sin (x - pi/2)) / cos (pi - x)] + [tan (x - 3pi/2) / - tan (pi + x)

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll re-write sin (x - pi/2) as -sin (pi/2 - x) = - cos x.

cos (pi - x) = cos pi*cos x + sin pi*sin x

But sin pi = 0 and cos pi = -1

cos (pi - x) = -cos x

The first term of the sum will become:

[sin (x - pi/2)) / cos (pi - x)] = - cos x/- cos  = 1

We'll re-write the numerator of the second term:

tan (x - 3pi/2) = - tan x

The denominator of the 2nd term is:

-tan(pi+x) = -(tan pi + tan x)/(1 - tanpi*tan x)

But tan pi = 0

-tan(pi+x) = -tan x/1

-tan(pi+x) = -tan x

We'll re-write the second term:

[tan (x - 3pi/2) / - tan (pi + x) = -tan x/-tan x = 1

The sum of the terms will be:

[sin (x - pi/2)) / cos (pi - x)] + [tan (x - 3pi/2) / - tan (pi + x) = 1+1

[sin (x - pi/2)) / cos (pi - x)] + [tan (x - 3pi/2) / - tan (pi + x) = 2

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lochana2500 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

the above answer has a little mistake :

-----------------------------

tan (x - 3π/2) = - tan x   ✗

tan (x - 3π/2)  = - cotx   ✓

-----------------------------

sin(x-π/2) / cos(π -x)  ⇒ ❶

= -sin (π/2-x)/-cosx

= -cosx/-cosx

=1

tan(x-3π/2)/-tan(π+x) ⇒ ❷

= -tan(3π/2-x)/-tanx

= -cot/-tanx

= cot²x

Hence ❶ + ❷ = 1 + cot²x = cosec²x

 

 

 

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