tan(A-theta)=(tan A-tan B)/(1+tan A*tan B)

Prove that tan(A-B)=(tan A-tan B)/(1+tan A*tan B) , if A=60° & B=45°.

In a triangle ABC ,right angled at A ,the altitude through A and internal bisector of /_A have lengths 3 and 4 respectively. Find the length of median through A . (Given that tan(A+B)=(tan A+tan B)/(1-tan A tan B),tan(A-B)=(tan A-tan B)/(1+tan A*tan B))

For all values of A and Btan(A+B)=(tan A+tan B)/(1-tan A tan B) and tan(A-B)=(tan A-tan B)/(1+tan A tan B)

If A=60⁰ and B=30⁰ , verify that tan (A-B)=(tan A-tan B)/(1+tan A tan B)

In a triangle ABC , right angled at A ,the altitude through A and internal bisector of /_A have lengths 3 and 4 respectively. Find the length of median through A. Given that tan(A+B)=(tan A+tan B)/(1-tan A tan B),tan(A-B)=(tan A-tan B)/(1+tan A tan B)

By using above basic addition/ subtraction formulae, prove that (i) tan (A+B)=(tan A+tan B)/(1-tan A tan B) , (ii) tan (A-B)=(tan A-tanB)/(1+tan A tan B) (iii) sin2theta=2sintheta costheta , (iv) cos2theta=cos^(2)theta=1-2sin^(2)theta=2cos^(2)theta-1 (v) tan 2theta=(2 tan theta)/(1-tan^(2) theta)

By using above basic addition/ subtraction formulae, prove that (i) tan (A+B)=(tan A+tan B)/(1-tan A tan B) , (ii) tan (A-B)=(tan A-tanB)/(1+tan A tan B) (iii) sin2theta=2sintheta costheta , (iv) cos2theta=cos^(2)theta=1-2sin^(2)theta=2cos^(2)theta-1 (v) tan 2theta=(2 tan theta)/(1-tan^(2) theta)

By using above basic addition/ subtraction formulae, prove that (i) tan (A+B)=(tan A+tan B)/(1-tan A tan B) , (ii) tan (A-B)=(tan A-tanB)/(1+tan A tan B) (iii) sin2theta=2sintheta costheta , (iv) cos2theta=cos^(2)theta=1-2sin^(2)theta=2cos^(2)theta-1 (v) tan 2theta=(2 tan theta)/(1-tan^(2) theta)

If A+B=(pi)/(2)rArr Tan A Tan B=1 , rArr Tan(A-B)=(tan A-tan B)/(1+tan A tan B)=(tan A-tan B)/(2) , rArr tan A=tan B+2tan(A-B) , (tan40^(@)+2tan10^(@))*(tan70^(@)-Tan20^(@))=

If A+B=(pi)/(2)rArr Tan A Tan B=1 , rArr Tan(A-B)=(tan A-tan B)/(1+tan A tan B)=(tan A-tan B)/(2) , rArr tan A=tan B+2tan(A-B) , x= Tan""(5 pi)/(28)+2Tan""(pi)/(7) Then x