Math, asked by apu825, 1 year ago

please help me as fast as u can..
answer will be marked as brainlist

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Answers

Answered by Yazlin
0

[cosx cosecx-sinx secx]/[cosx+sinx]

[(cosx/sinx)-(sinx/cosx)]/[cosx+sinx][(cos^2x-sin^2x)/sinxcosx]/[cosx+sinx]

=[(cosx-sinx)(cosx+sinx)]/[(cosx+sinx)

(sinxcosx)

=[(cosx-sinx)/sinxcosx]

=(cosx/sinxcosx)-(sinx/sinxcosx)

=(1/sinx)-(1/cosx)

=cosecx-secx


Yazlin: tnx for the brainliest
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