sec theta +tan theta =4 to 51sin theta +34 cos theta
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Sec theta=1/cos theta and tan theta=sin theta/cos theta therefore rewriting the equation we have:1/costheta+sin theta/cos theta =4
1/cos theta(1+sin theta) =4 therefore
Cos theta=1+sin theta /4 and sin theta=4cos theta-1.
Therefore the answer is:51*(4cos theta-1)+34*((1+sin theta) /4)
204cos theta-51+(34+34sin theta) /4
1/cos theta(1+sin theta) =4 therefore
Cos theta=1+sin theta /4 and sin theta=4cos theta-1.
Therefore the answer is:51*(4cos theta-1)+34*((1+sin theta) /4)
204cos theta-51+(34+34sin theta) /4
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Hey i can tell you the approach hope you can easily solve:
#Convert sec∅ and tan∅ into sin-cos form you will get 1+sin∅=4cos∅.Square both sides you will get 1+sin^2∅+2sin∅=16cos^2∅.
#Put 1-sin^2∅ at the place of cos^2∅.It will convert into quadratic equation in sin∅.
#Further on solving you will get ∅.Now you can put this ∅ in the above expression to get the answer.
#Convert sec∅ and tan∅ into sin-cos form you will get 1+sin∅=4cos∅.Square both sides you will get 1+sin^2∅+2sin∅=16cos^2∅.
#Put 1-sin^2∅ at the place of cos^2∅.It will convert into quadratic equation in sin∅.
#Further on solving you will get ∅.Now you can put this ∅ in the above expression to get the answer.
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