trig functions (1 Viewer)

[jemsta]

hey ppl need help with a few questions
1) prove
tan2xcotx=1+sec2x
2) prove
cos^2acos^2b-sin^2asin^2b= 1/2(cos2a+cos2b)
thanks

 

[香港!]

hey ppl need help with a few questions
1) prove
tan2xcotx=1+sec2x
2) prove
cos^2acos^2b-sin^2asin^2b= 1/2(cos2a+cos2b)
thanks

1)
tan2xcotx=2tanx\(1-tan²x)tanx=2\(1-tan²x)=2cos²x\(cos²x-sin²x)=2cos²x\cos2x
1+sec2x=1+1\(2cos²x-1)=(2cos²x)\(2cos²x-1)=2cos²x\cos2x
2)cos²a cos²b-sin²asin²b=(cosacosb-sinasinb)(cosacosb+sinasinb)
=cos(a+b)cos(a-b)=(1\2)[2cos(a+b)cos(a-b)]=(1\2)(cos2a+cos2b)

 

[KFunk]

hey ppl need help with a few questions
1) prove
tan2xcotx=1+sec2x

LHS = 2tanx/(1-tan²x).cotx

= 2/(1 - tan²x)

= 2cos²x/(cos²x - sin²x) (multiplying through by cos²x)

= (cos²x + cox²x)/(cos2x)

= (cos²x - sin²x + 1)/cos2x .... (since cos²x = 1 - sin²x)

= (cos2x + 1)/cos2x = 1 + sec2x

 

[icycloud]

Another method for (2):

LHS = cos²a cos²b-sin²a sin²b
= cos²a cos²b-(1-cos²a)(1-cos²b)
= cos²a cos²b-(1-cos²b-cos²a+cos²a cos²b)
= cos²a+cos²b-1

RHS = 1/2(cos(2a)+cos(2b))
= 1/2(2cos²a-1+2cos²b-1)
= cos²a-1/2+cos²b-1/2
= cos²a+cos²b-1
= LHS #