Question

show that tan Cube theta -1upon tan theta - 1 =sec square theta + tan stheta
please answer this question ​

Answer 1

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Answer 2

Solution:- From LHS

tan³¢ - 1/tan¢ - 1
or, tan³¢ - 1³ /tan¢ - 1 , Using identity, a³ - b³ = (a -b)(a²-b² + ab)
or, (tan¢ - 1)(tan²¢ - 1 + tan¢) / ( tan¢ - 1)
or, tan²¢ - 1 + tan¢
or, sec²¢ + tan¢. , {tan²¢ - 1 = sec²¢)

hence, prooved , sec²¢ + tan¢ RHS

LHS = RHS prooved