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# Trig identities

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How would one solve this?

[(1+tan(x))/sin(x)]-sec(x)=csc(x)

May 25, 2018

#1
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Verify the following identity:

(tan(x) + 1)/sin(x) - sec(x) = csc(x)

Put (tan(x) + 1)/sin(x) - sec(x) over the common denominator sin(x): (tan(x) + 1)/sin(x) - sec(x) = (1 - sec(x) sin(x) + tan(x))/sin(x):

(1 - sec(x) sin(x) + tan(x))/sin(x) = ^?csc(x)

Multiply both sides by sin(x):

1 - sec(x) sin(x) + tan(x) = ^?csc(x) sin(x)

Write cosecant as 1/sine, secant as 1/cosine and tangent as sine/cosine:

1 - 1/cos(x) sin(x) + sin(x)/cos(x) = ^?1/sin(x) sin(x)

1 - (1/cos(x)) sin(x) + (sin(x)/cos(x)) = 1:

1 = ^?(1/sin(x)) sin(x)

(1/sin(x)) sin(x) = 1:

1 = ^?1

The left hand side and right hand side are identical:

(identity has been verified)

May 25, 2018
#2
+102383
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(1 + tan x  )  / sin x  - sec  x = csc x

(1 + tan x) / sin x  -    1 / cos x

(1 + sin x / cos x) /  sin x  - 1 / cos x

1/sin x + sinx / [ cos x sin x]  - 1/cos x

1 / sin x  + sin x / [ cos x sin x ]  - sin x / [ sinx cos x]

1 / sin x  + sin x [ sin x cos x] -  sin x / [ sin x cos x ]

1 / sin x   =    csc x

May 25, 2018