can someone pls help me in solve these question plssssssss plssss
Answers
Answer:
Step-by-step explanation:
1) ) cosec 3x = cot 30° + cot 60°1 + cot 30° cot 60°
cot 30° = √3 ; cot 60° = 13√
cot 30° + cot 60° 1 + cot 30° cot 60° = 3√+13√1+3√×13√ = (3+1)3√2 = 43√ × 12= 23√
So, cosec 3x = 23√
cosec 3x = cosec 60°
∴ 3x = 60° or x = 20°
2) Let the three numbers be a-d, a, a+d.
It is given that the sum of these number is 21.
a-d+a+a+d=21a−d+a+a+d=21
3a=213a=21
a=7a=7
The value of a is 7.
The product of their extremes is 45.
(a-d)(a+d)=45(a−d)(a+d)=45I
a^2-d^2=45a
2
−d
2
=45
7^2-d^2=457
2
−d
2
=45
49-45=d^249−45=d
2
4=d^24=d
2
d=\pm 2d=±2
The value of d is either 2 or -2.
Therefore the three numbers are either 5,7,9 or 9,7,5.
3) a+2d +
a+8d equal to8 given..
now 11/2(2à +(11-1)d) ll be equal to 11/2(2a+10d)..from above we know that 2a+10d is equal to 8 so ans ll be 11/2×8 =44
4) This I don't know
5) tan a/(1-cot a) +cot a/(1-tan a)
=(sin a/,cos a) /(1-cos a/sin a) + (cos a/sin a) /(1-sin a/cos a)
=sin ^2 a/cosa(sina - cosa) +cos^2 a/sina (cosa-sina)
=sin^2a/cosa(sina-cosa) - cos^2a/sina (sina-cosa)
=(sin^3a-cos^3a)/sina.cosa(sina-cosa)
=(sina-cosa)(sin^2a+cos^2a+sinacosa)/sina.cosa(sina-cosa)
=(1+sinacosa)/sina.cosa
=(1/sinacosa)+1
=1+seca.coseca.