If $$cos$$θ$$1=2cos$$θ$$2$$ and $$Tan($$θ$$1$$−θ$$22)Tan($$θ$$1+$$θ$$22)=$$−pq, then $$p+q=$$

Question

If $$cos$$θ$$1=2cos$$θ$$2$$ and $$Tan($$θ$$1$$−θ$$22)Tan($$θ$$1+$$θ$$22)=$$−pq,  then   $$p+q=$$

Infinity Learn · Accepted Answer

We  have    $$tan$$$$($$θ$$1$$−θ$$22)$$ $$tan$$$$($$θ$$1+$$θ$$22)=$$ $$2sin($$θ$$1-$$θ$$22)$$$$sin$$$$($$θ$$1+$$θ$$22)2cos($$θ$$1-$$θ$$22)$$$$cos$$$$($$θ$$1+$$θ$$22)$$                                                                          $$=$$$$cos$$θ$$2$$−$$cos$$θ$$1cos$$θ$$1+$$$$cos$$θ$$2=$$$$cos$$θ$$2$$−$$2cos$$θ$$22cos$$θ$$2+$$$$cos$$θ$$2$$                                                                   ⇒$$p=1$$   and   $$q=3$$                                                                     −$$pq=$$−$$13$$                                                     ∴$$p+q=4$$

Trigonometric equations

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If $\cos θ_{1} = 2 \cos θ_{2}$ and $T a n (\frac{θ_{1} - θ_{2}}{2}) T a n (\frac{θ_{1} + θ_{2}}{2}) = \frac{- p}{q}, t h e n p + q =$

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Trigonometric equations

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If cosθ1=2cosθ2 and Tan(θ1−θ22)Tan(θ1+θ22)=−pq, then p+q=

We have tan(θ1−θ22) tan(θ1+θ22)= 2sin(θ1-θ22)sin(θ1+θ22)2cos(θ1-θ22)cos(θ1+θ22) =cosθ2−cosθ1cosθ1+cosθ2=cosθ2−2cosθ22cosθ2+cosθ2 ⇒p=1 and q=3 −pq=−13 ∴p+q=4

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If $\cos θ_{1} = 2 \cos θ_{2}$ and $T a n (\frac{θ_{1} - θ_{2}}{2}) T a n (\frac{θ_{1} + θ_{2}}{2}) = \frac{- p}{q}, t h e n p + q =$