Question

1 - cos 0
( (cosec 0 - cot 0)2 =
1 + cos 0​

Answer 1

Step-by-step explanation:

Your question is hundred percent wrong

Answer 2

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

CORRECT QUESTION:

$\bullet \ \ \sf (cosec \theta-cot \theta)^2=\dfrac{1-cos \theta}{1+cos \theta}$

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

LHS:

$\to \sf (cosec \theta-cot \theta)^2$

$\to \sf \left (\dfrac{1}{sin \theta} -\dfrac{cos \theta}{sin \theta} \right)^2$

$\to \sf \left (\dfrac{1-cos\theta}{sin \theta} \right)^2$

$\to \sf \left(\dfrac{(1-cos\theta)^2}{ \bold{sin^2 \theta}} \right)$

Simplify the boldened part by substituting sin²θ = 1 - cos²θ.

$\to \sf \left(\dfrac{(1-cos\theta)^2}{\bold{1-cos^2 \theta}} \right)$

Simplify the boldened part by using (a - b)² = a² - b².

$\to \sf \left(\dfrac{(1-cos\theta)(1-cos\theta)}{(1-cos \theta)(1+cos \theta)} \right)$

$\to \sf{\red{ \left(\dfrac{(1-cos\theta)}{(1+cos \theta)} \right)}} \bigstar$

LHS=RHS.

HENCE PROVED.

FUNDAMENTAL TRIGONOMETRIC IDENTITIES:

$\boxed{\substack{\displaystyle \sf sin^2 \theta+cos^2 \theta = 1 \\\\ \displaystyle \sf 1+cot^2 \theta=cosec^2 \theta \\\\ \displaystyle \sf 1+tan^2 \theta=sec^2 \theta}}$

TRIGONOMETRIC TABLE:

$\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}$

You can find cot, cosec and sec by reciprocating tan, sin and cos.