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abhiraj8:
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Sino=ksin(o+fi) show that tan(o+fi) =sin(fi) /cos(fi)-k ( o=Thita)
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Please find below the solution to the asked query:
If you observe we don't need θ in the final prrof. Given that,sinθ=ksin(θ+ϕ)using identity sin(A+B)=sinA.cosB+cosA.sinB, we getsinθ=k(sinθ.cosϕ+cosθ.sinϕ)Dividing each term by sinθ we get,1=kcosϕ+kcosθsinθ.sinϕwe know that cosθsinθ=1tanθ1=kcosϕ+ktanθ.sinϕ⇒ktanθ.sinϕ=1−kcosϕ⇒1tanθ=1−kcosϕksinϕ⇒tanθ=ksinϕ1−kcosϕNow tan(θ+ϕ)=tanθ+tanϕ1−tanθ.tanϕPutting the value of tanθ, we get, tan(θ+ϕ)=
Please find below the solution to the asked query:
If you observe we don't need θ in the final prrof. Given that,sinθ=ksin(θ+ϕ)using identity sin(A+B)=sinA.cosB+cosA.sinB, we getsinθ=k(sinθ.cosϕ+cosθ.sinϕ)Dividing each term by sinθ we get,1=kcosϕ+kcosθsinθ.sinϕwe know that cosθsinθ=1tanθ1=kcosϕ+ktanθ.sinϕ⇒ktanθ.sinϕ=1−kcosϕ⇒1tanθ=1−kcosϕksinϕ⇒tanθ=ksinϕ1−kcosϕNow tan(θ+ϕ)=tanθ+tanϕ1−tanθ.tanϕPutting the value of tanθ, we get, tan(θ+ϕ)=
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