Prove: `sin^2θ`+`cos^2θ`=1

Prove that;- sin^6θ+cos ^6θ=1−3sin^2θcos ^2θ

The values of θ lying between 0 and π/2 and satisfying the equation ∣ 1+sin ^2θ sin^2θ sin^2θ cos^2θ 1+cos^2θ cos^2θ 4sin4θ 4sin4θ 1+4sin4θ ∣ =0 are

Prove that sin(θ+30°)+cos(θ+60°)= cos θ.

Prove that (sec^ 2θ−1)(cosec^ 2θ−1)=1

tan θ.sin^2 θ=

If cosα+cosβ= 3/2 ​ and sinα+sinβ= 1/2 ​ and θ is the arithmetic mean of α and β, then sin2θ+cos2θ is equal to

If cosα+cosβ= 3/2 ​ and sinα+sinβ= 1/2 ​ and θ is the arithmetic mean of α and β, then sin2θ+cos2θ is equal to

The number of all possible values of θ where 0 sin3θ = 2cos((3θ)/y )+ 2sin((3θ)/z) (xyz)sin3θ = (y + 2z)cos3θ + ysin3θ has a solution (x_0,y_0,z_0) with y_0 ,z_0 ≠ 0, is

If 32tan ^8 θ=2cos^2 α−3cosα and 3cos2θ=1 then the general value of α is

Prove that : (sin A)/ (1 + cos A) + (1 + cos A)/ (sin A) = 2 cosec A