Trigonometric Ratios of Compound Angles

Asked by  | 23rd Mar, 2008, 09:29: PM

Expert Answer:

1/ tan3A-tanA    -        1/cot3A-cotA  =  cot2A 

1/ 1/ cot3A - 1/ cotA  -  1/ cot3A - cotA

cotA cot3A/ cotA - cot3A   -  1/ cotA - cot3A

1 + cotA cot3A / cotA - cot3A

(1 + 1/ tanA.tan3A )/ (1/ tanA - 1/ tan3A)

(tanA tan3A +1)/ (tan3A - tanA )  

{ tanA ( 3 tanA - tan3A/ 1-3tan2A) + 1} / {( 3 tanA - tan3A)/1 - 3tan2A} - tanA 

                                                                                   tan3A = 3 tanA - tan3A/ 1-3tan2A

(1 - tan 4 A)/ (2 tan A + 2tan3A)  = (1 - tan2A)/ 2 tanA

 = ( 1 - tan2A)( 1+tan2A)/ 2tanA ( 1+tan2A) =  ( 1 - tan2A)/ 2tanA

 = (cotA - 1)/2 cot2A  = cot2A

                               Proved 

Answered by  | 11th Jun, 2008, 05:05: PM

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