if 1/cos-cos=a^3 and 1/sin-sin=b^3 show that tan=a/b

Question

if it is given that tan(a-b)=tanA-tanB/1+tanAtanB AND tanP-1/1+tanP=tan195

find the value of P

Answer 1

(1) sec(x)-cos(x)=a³, csc(x)-sin(x)=b³;

(1-cos²(x))/cos(x)=a³, (1-sin²(x))/sin(x)=b³;

sin²(x)/cos(x)=a³; cos²(x)/sin(x);

(sin²(x)/cos(x))(cos²(x)/sin(x))=a³/b³;sin³(x)/cos³(x)=a³/b³.

Take cube root of each side: sin(x)/cos(x)=tan(x)=a/b QED

(2) (tan(P)-1)/(1+tan(P))=tan(195);

tan(P)-1=tan(195)+tan(P)tan(195);

tan(P)-tan(P)tan(195)=1+tan(195);

tan(P)=(1+tan(195))/(1-tan(195))=

(tan(45)+tan(195))/(1-tan(45)tan(195)=

tan(45+195); so P=240°.

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