Question

Q.Prove that Sin²∅Cot²∅+Sin²∅=1​

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Answer 1

Step-by-step explanation:

Let ABC be a right angled triangle right angled at B.

Let angle C ie angle ACB = theeta.

Now by Pythagoras theorem

AC^2= AB^2+BC^2 ……(1)

We know that sin theeta = opposite side/ hypotenuse.

So sin theeta = AB/AC. Similarly

cos theeta= adjacent side/ hypotenuse

So cos theeta = BC/AC

Now sin^2 theeta + cos^2 theeta

= (AB/AC)^2 + (BC/AC)^2

= AB^2/AC^2 + BC^2/AC^2

= (AB^2+BC^2)/AC^2

= AC^2/AC^2 = 1 ( by using (1) )

Answer 2

Step-by-step explanation:

given :

• Q.Prove that Sin²∅Cot²∅+Sin²∅=1

to find :

• Sin²∅Cot²∅+Sin²∅=1

solution :

sin = Sin²0. cos²Ø / sin²Ø + sin²0

sin = sin²0.cos²0/sin²0 + sin²0

sin = cos²0 + sin ²0

sin = 1

hence, sin = 1