Question

Pls Answer Fast it is Attachment I will mark as Brainliest

Answer 1

$\mathcal \red \bigstar{\underline{\purple{SOLUTION:-}}}$

REFER TO THE ATTACHMENT ABOVE:-

(For urgent I solved it in Note book)........

@Vyomsingh

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

$\mathcal \red \bigstar{\underline{\purple{FORMULA \:USED:-}}}$

(a+b)^3=a^3+b^3+3a^2b+3ab^2

(a-b)^3=a^3-b^3-3a^2b+3ab^2

tan a=1/cota

or

cot a=1/tan a