30 - Trigonometry Questions Answers


what is the angular difference between the +j and -j ?
OR
+4j and -4j ?
180 degree
sin‾¹(sin2) + sin‾¹(sin4) + sin‾¹(sin6) = ?
sin‾¹(sin2) + sin‾¹(sin4) + sin‾¹(sin6)
= (π-2) + (π-4) + (6-2π)
= 0
Solve
tanx-3cotx=2tan3x(0<x<360)
cosx-sinx=cosy-siny
cos2xcotx+1=cos2x+cotx
tanx - 3/tanx = 2[3tanx-tan3x]/[1-3tan2x]
so [tan2x - 3]/tanx = 2tanx[3-tan2x]/[1-3tan2x]
so [tan2x - 3][1-3tan2x] = 2tan2x[3-tan2x]
so [tan2x - 3][1-3tan2x] + 2tan2x[tan2x-3] = 0
so [tan2x - 3][1-tan2x] = 0
so tanx = 1, -1, √3, -√3
now check the values satisfying last eq.
then put these values in 2nd eq. to get answer.
Find the value :-
(tan69 +tan66) / (1 - tan69.tan66)
use formula of tan(A+B)
Find the max. and min. value of
y = 7cosA + 24sinA
7cosA + 24sinA
= 25[7cosA/25 + 24sinA/25] = 25[cosAcosB + sinAsinB] where tanB = 24/7
= 25cos[A-B] max and min of cos[A-B] = 1 and -1
so max. = 25 and min. = -25
Prove that :-
tanA + 2tan2A + 4tan4A + 8cot8A = cosecA / secA
We have to prove
tanA + 2tan2A + 4tan4A + 8cot8A = cosecA / secA
or tanA + 2tan2A + 4tan4A + 8cot8A = cotA
or 2tan2A + 4tan4A + 8cot8A = cotA - tanA
or tan2A + 2tan4A + 4cot8A = [1-tan2A]/2tanA
or tan2A + 2tan4A + 4cot8A = cot2A
or 2tan4A + 4cot8A = cot2A - tan2A
or tan4A + 2cot8A = [1-tan22A]/2tan2A
or tan4A + 2cot8A = cot4A
or 2cot8A = cot4A-tan4A
or cot8A = [1-tan24A]/2tan4A
it is true so proved
Prove that :-
sin(A+B+C+D) + sin(A+B-C-D) + sin(A+B-C+D) + sin(A+B+C-D) = 4sin(A+B).cosC.cosD
sin(A+B+C+D) = sin(A+B)[cosCcosD-sinCsinD]+cos(A+B)[sinCcosD+cosCsinD]
similarly solve all the terms and add all these terms for getting answer
((1(secA.secA-cosA.cosA))-(1/(cosecA.cosecA-sinA.sinA))).sinA.sinA.cosA.cosA=(1-cosA.cosA.sinA.sinA)/(2+cosA.cosA.sinA.sinA)
Submit the correct question, In first bracket / will be there so
((1/(secA.secA-cosA.cosA))-(1/(cosecA.cosecA-sinA.sinA))).sinA.sinA.cosA.cosA
= [cos2A/sin2A(1+cos2A) - sin2A/cos2A(1+sin2A)]sin2Acos2A
= [cos4A(1+sin2A) - sin4A(1+cos2A)] / [(1+sin2A)(1+cos2A)]
now solve