" (b) "tan alpha=(16)/(63)

If sin (alpha + beta) = 4/5 , sin (alpha -beta) = (5)/(13), alpha + beta , alpha - beta being acute angles prove that tan 2 alpha = (63)/(16).

If a tan alpha+b tan beta=(a+b) tan ((alpha+beta)/2) , where alpha ne beta , prove that : a cos beta= b cos alpha .

If alpha ne beta " and " a tan alpha + b tan beta = (a + b) "tan" (alpha + beta)/2,"show that" (cos alpha)/(cos beta) = a/b.

If tan alpha=(1-cos beta)/(sin beta), then a )tan3 alpha=tan2 beta(b)tan2 alpha=tan beta(c)tan2 beta=tan alpha(d) none of these

Find the values of cos 2A, A lies in the first quadrant, when tan A = (16)/(63) .

tan alpha+2tan2 alpha+4tan4 alpha+8tan8 alpha+16tan16 alpha+32cot32 alpha. is equal to: (1)cot alpha(2)tan alpha.(3)cos alpha(4)sin alpha

If a tan alpha+b tan beta=(a+b)tan((alpha+beta)/(2)) where alpha!=beta ,prove that a cos beta=b cos alpha

(sin(alpha+beta))/(sin(alpha-beta))=(a+b)/(a-b) then (tan alpha)/(tan beta)=

If alpha=(pi)/(14), then the value of (tan alpha tan2 alpha+tan2 alpha tan4 alpha+tan4 alpha tan alpha) is 1(b)1/2(c)2(d)1/3