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iragen [17]
3 years ago
6

Show that: (sec theta - cosec theta) (1 + tan theta + cot theta) = sec theta tan theta - cosec theta cot theta)

Mathematics
1 answer:
kvv77 [185]3 years ago
4 0

Answer:

see derivation below

Step-by-step explanation:

Show that:

( sec(t) - cosec(t) ) ( 1 + tan(t) + cot(t) ) =

sec(t) tan(t) - cosec(t) cot(t)

Some trigonometric definitions used:

tan(t) = sin(t)/cos(t)

cot(t) = cos(t)/sin(t)

sec(t) = 1/cos(t)

csc(t) = 1/sin(t)

some trigonometric identities used:

sin^2(t) + cos^2(t) = 1 ......................(1)

rewrite left-hand side in terms of sine and cosine

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

Simplify using common denominator sin(t)cos(t)

= ( (sin(t) - cos(t))/(sin(t)*cos(t)) ) * ( ( sin(t)cos(t) + sin^2(t) + cos^2(t)) / ( sin(t)cos(t) ) )

= ( sin(t) -cos(t) ) * (1 + sin(t)cos(t) ) / ( sin^2(t) cos^2(t) )   ...... using (1)

Expand by multiplication

= ( sin(t) -cos(t) + sin^2(t)cos(t) - sin(t)cos^2(t) ) / ( sin^2(t) cos^2(t) )

Rearrange by factoring out sin(t) and cos(t) in numerator

= ( sin(t) (1-cos^2(t) - cos(t)(1-sin^2(t) )  / ( sin^2(t) cos^2(t) )

= ( sin^3(t) - cos^3(t) ) /( sin^2(t) cos^2(t) )   .........................using (1)

Cancel common factors

= sin(t)/(cos^2(t)) - cos(t)/(sin^2(t))

Rewrite using trigonometric definitions

= sec(t)tan(t) - csc(t)cot(t)   as in Right-Hand Side

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Answer:

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Step-by-step explanation:

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In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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