Prove that tan^3 theta-1 by tan theta-1=sec^2 theta+tan theta?

basic trigonometry or:basic trigonometryor:(Tan^3\u03b8-1)/Tan\u03b8 =Sec^2\u03b8+Tan\u03b8R.H.S=Sec^2\u03b8+Tan\u03b8 =(1/Cos^2\u03b8) + (

basic trigonometry

or:basic trigonometry

or:(Tan^3\u03b8-1)/Tan\u03b8 =Sec^2\u03b8+Tan\u03b8R.H.S=Sec^2\u03b8+Tan\u03b8 =(1/Cos^2\u03b8) + (Sin\u03b8/Cos\u03b8) =(1+Cos\u03b8.Sin\u03b8)/Cos^2\u03b8 =(1/Cos^2\u03b8) + (Cos\u03b8.Sin\u03b8)/Cos^2\u03b8 =Sec^2\u03b8+(Sin\u03b8/Cos\u03b8) =Sec^2\u03b8+Tan\u03b8Hence Proved

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