![SOLVED: sin(θ) = sin(θ + 27°) = sin(θ + π/2) = cos(θ) = sin(θ + 22°) = sin(θ) cos(θ) + cos(θ) sin(θ) = 1 cos(θ) = cos(θ) cos(θ + 2π) = cos(θ) SOLVED: sin(θ) = sin(θ + 27°) = sin(θ + π/2) = cos(θ) = sin(θ + 22°) = sin(θ) cos(θ) + cos(θ) sin(θ) = 1 cos(θ) = cos(θ) cos(θ + 2π) = cos(θ)](https://cdn.numerade.com/ask_images/ea3e561f91574a5dbca5d3772fbcb8c7.jpg)
SOLVED: sin(θ) = sin(θ + 27°) = sin(θ + π/2) = cos(θ) = sin(θ + 22°) = sin(θ) cos(θ) + cos(θ) sin(θ) = 1 cos(θ) = cos(θ) cos(θ + 2π) = cos(θ)
File:The cotangent function cot(z) plotted in the complex plane from -2-2i to 2+2i.svg - Wikimedia Commons
![cosec x= cosec y cosec z + cot y cot z , then prove that cosec y = cosecx - Maths - Trigonometric Functions - 14117073 | Meritnation.com cosec x= cosec y cosec z + cot y cot z , then prove that cosec y = cosecx - Maths - Trigonometric Functions - 14117073 | Meritnation.com](https://s3mn.mnimgs.com/img/shared/content_ck_images/ck_5f55ee0b5a95f.jpg)
cosec x= cosec y cosec z + cot y cot z , then prove that cosec y = cosecx - Maths - Trigonometric Functions - 14117073 | Meritnation.com
![complex analysis - How to show that $\displaystyle{\cot z-\frac{1}{z}}$ is bounded on the given circle - Mathematics Stack Exchange complex analysis - How to show that $\displaystyle{\cot z-\frac{1}{z}}$ is bounded on the given circle - Mathematics Stack Exchange](https://i.stack.imgur.com/CrVp1.png)
complex analysis - How to show that $\displaystyle{\cot z-\frac{1}{z}}$ is bounded on the given circle - Mathematics Stack Exchange
![If x+y+z=π prove the trigonometric identity cot x/2 + cot y/2 + cot z/2 =cot x/2 .cot y/2 .cot z/2 . - Brainly.in If x+y+z=π prove the trigonometric identity cot x/2 + cot y/2 + cot z/2 =cot x/2 .cot y/2 .cot z/2 . - Brainly.in](https://hi-static.z-dn.net/files/d60/8656d1fc377e30c50669e20f73c8fd9f.jpg)