Question

koi iss question ko solve kardo please...... ​

Answer 1

Answer:

$\theta = 2n\pi +- \frac{\pi}{3}$

$\pink{solution}: -$

We know that , tan¢ = sin¢ /cos¢

and ,cot¢ = cos¢ /sin¢ ,cosec¢ = 1/sin¢

=>3(tan¢) + cot¢ = 5cosec¢

=> [3(sin¢/cos¢ ) + cos¢/sin¢] = 5/sin¢

=> [3sin²¢+cos²¢/sin¢*cos¢ ] = 5/sin¢

=> 3(1-cos²¢) + cos²¢ ] = 5cos¢[ •°•sin²¢ = 1-cos²¢]

=> 3- 3cos²¢ + cos²¢ - 5cos¢ =0

=> 2cos²¢ +5cos¢ -3 = 0

=> 2cos²¢ + 6cos¢ -cos¢ - 3 = 0

=> 2cos¢ ( cos¢ + 3) - 1(cos¢ +3) = 9

=>( 2cos¢ -1) (cos¢ +3) = 0

cos¢ = 1/2 or cos¢ = -3 ( neglected) cause range of cos¢ is [-1,1]

so, cos¢ = 1/2 => cos¢ = cosπ/3 =>¢ = π/3

¢ = 2nπ +- π/3 using general solution of trigonometry

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$\boxed{hope \: its \: helpful}$

Answer 2

Answer:

θ=2nπ+−

3

π

\pink{solution}: -solution:−

We know that , tan¢ = sin¢ /cos¢

and ,cot¢ = cos¢ /sin¢ ,cosec¢ = 1/sin¢

=>3(tan¢) + cot¢ = 5cosec¢

=> [3(sin¢/cos¢ ) + cos¢/sin¢] = 5/sin¢

=> [3sin²¢+cos²¢/sin¢*cos¢ ] = 5/sin¢

=> 3(1-cos²¢) + cos²¢ ] = 5cos¢[ •°•sin²¢ = 1-cos²¢]

=> 3- 3cos²¢ + cos²¢ - 5cos¢ =0

=> 2cos²¢ +5cos¢ -3 = 0

=> 2cos²¢ + 6cos¢ -cos¢ - 3 = 0

=> 2cos¢ ( cos¢ + 3) - 1(cos¢ +3) = 9

=>( 2cos¢ -1) (cos¢ +3) = 0

cos¢ = 1/2 or cos¢ = -3 ( neglected) cause range of cos¢ is [-1,1]

so, cos¢ = 1/2 => cos¢ = cosπ/3 =>¢ = π/3

¢ = 2nπ +- π/3 using general solution of trigonometry

___________________________

\boxed{hope \: its \: helpful}

hopeitshelpful