(1+sinθ) (1-sinθ)

(sinθ - 2sin^3θ)/(2cos^3θ - cosθ) = tanθ

Prove the following trigonometric identities: sinθ/(1+cosθ) + (1+cosθ)/sinθ = 2/sinθ

The value of ((sinθ -cosθ )(1 + tanθ + cotθ ))/(1+sinθ cosθ ) = ? ((sinθ -cosθ )(1 + tanθ + cotθ ))/(1+sinθ cosθ ) का मान है :

If (sinθ/(1+ cosθ)) + ((1 +cosθ)/sinθ) = 4/sqrt3 0^circ lt θ lt 90^circ , then the value of (tanθ + secθ)^(-1) is : यदि (sinθ/(1+ cosθ)) + ((1 +cosθ)/sinθ) = 4/sqrt3 0^circ lt θ lt 90^circ ,, तो (tanθ + secθ)^(-1) का मान है :

The value of ( 1 + cotθ – cosecθ ) ( 1 + cosθ + sinθ )secθ = ? ( 1 + cotθ – cosecθ ) ( 1 + cosθ + sinθ )secθ का मान है :

If 5sin θ – 4cos θ = 0, 0^circ lt θ lt 90^circ , then the value of (5sinθ− 2cosθ)/ (5 sin θ + 3cosθ) is: यदि 5sin θ – 4cos θ = 0, 0^circ lt θ lt 90^circ है, तो (5sinθ− 2cosθ)/ (5 sin θ + 3cosθ) का मान है:

x = a(θ - sinθ), y = a(1 + cosθ) Find dy/dx

If ((sinθ−cosecθ)(cosθ−secθ))/(tan^(2)θ -sin^(2)θ) = r^3 , then r=? यदि ((sinθ−cosecθ)(cosθ−secθ))/(tan^(2)θ -sin^(2)θ) = r^3 है तो r बराबर हैः

Point (1,1), (1,-1), (-1, 1), (-1,-1)

If [{:(1, 1), (0,1):}]*[{:(1, 2), (0,1):}]*[{:(1, 3), (0,1):}]cdotcdotcdot[{:(1, n-1), (0,1):}] = [{:(1, 78), (0,1):}] , then the inverse of [{:(1, n), (0,1):}] is