find the value of the following :- sin(300) cosec(1050) - tan(-120)
@neer2890 @Kainui @IMStuck @Koikkara
Try your best first to attempt it.
sin(300) = -1/2 ..... is it ok
you can use the same method used by earlier to find their values.
is it ok ?
sin(300)=sin(90*3+30) (90*3+30) indicates 4th quadrant in which sin theta is negative. so ans is negative. also, multiple of 90 is odd, so sin changes to cos sin(300)=- cos 30
sin(300)=- cos 30 how ?
it must be cosec I think
(90*3+30) is in 4th quadrant , so now we have to chk for the trignometric fxn associated with it. which is sin theta. and sin is -ve in 4th quadrant. so a -ve sign should be associated with the answer. also, multiple of 90 is odd, so sin changes to cos. so sin(300)=-cos30
got it ?
ya
also, cosec(1050)=cosec(90*11+60) (90*11+60) is in 4th quadrant in which cosec is negative, so a -ve sign is associated with the answer. and multiple of 90 is odd , so cosec changes to sec. so cosec(1050)=-sec60=-2
tan (-120)=-tan(120)=-tan(90*1+30) (90*1+30) is in 2nd quadrant in which tan is -ve so a -ve sign is associated with the answer. also, multiple of 90 is odd , so tan changes to cot. so -tan(120)=-(-cot30)=cot30= sqrt(3)
\[\sin 300 cosec 1050- \tan(-120)= -\frac{ \sqrt{3} }{ 2 }*(-2)-\sqrt{3}=\sqrt{3}-\sqrt{3}=0\]
How to judge if 90*678 is in which quadrant ? @neer2890
and also when its 90*1159
we know we have 4 quadrants,check for multiple of 4
how ?
ok. for e.g., \[90*4+ \Theta\] is in 1st quadrant. for 90*678=(90*(169*4+2))
so 90*169*4 is 4th quadrant and add 90*2 . so 90*678=2nd quadrant
-tan(120)= -tan(90*1+30 ) .............. after that ?
@neer2890
= - cot 30 = - sqrt 3
as cot 30 = sqrt 3
ok.(90*1+30) is 2nd quadrant in which tan is negative. so extra negative sign is compulsory with answer. als0, multiple of 90 is odd, so tan changes to cot -tan(90*1+30)=-(-cot30)=cot30
in -(-cot 30), 1st minus sign is of -tan(120) and 2nd minus is because tan is -ve in 2nd quadrant.
No finally I got it Bhaiya
Yu Hu Thank You
you're welcome...:)
@neer2890 are you in IIT ?
no.. but i studied from IIT professors.
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