Question

if cosec thita =2 and cot thita = √3P. where thita is an acute angle, then wht is the value of P

Answer 1

Answer:

±1

Step-by-step explanation:

Given---> Cosecθ=2 ,Cotθ=(√3)p

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To find--->Value of p i.e. p=?

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Solution----> We know that

------------ 1+Cot²θ=Cosec²θ

Now Cosecθ=2 ,Cotθ=(√3)p

Cosec²θ - Cot²θ =(2)² - (√3 p)²

=> 1 = 4 - 3p²

=> 3p² =4-1

=> 3p²= 3

=> p²=3/3

=> p²=1

=> p=±1

Additional information---->

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1)Sin²θ+Cos²θ=1

2)1+tan²θ=Sec²θ

3)tan²θ=Sec²θ-1

4)1=Sec²θ-tan²θ

5)1-Cos²θ=Sin²θ

6)1-Sin²θ=Cos²θ

Answer 2

$\huge\mathfrak\red{Answer}$

$\csc \alpha = 2..........(1) \\ \cot \alpha = p \sqrt{3} ......(2) \\ \\ take \: square \: on \: (2) \: \\ \\ {cot}^{2} \alpha = 3 {p}^{2} \\ \\ we \: know \: that \: 1 + {cot}^{2} \alpha = {csc}^{2} \alpha \\ \\ { \csc}^{2} \alpha - 1 = 3 {p}^{2} \\ from \: (1) \: we \: know \: that \: csc \alpha = 2 \\ \\ {(2)}^{2} -\: 1 = 3 {p}^{2} \\ 3 = 3 {p}^{2} \\ {p}^{2} = 1 \\ p = - 1 \: or \: + 1 \\ \\ as \: you \: mentioned \: that \: \alpha \: is \: an \: acute \\ \: angle \: and \: we \: know \: that \: cot \alpha \: is \\ positive \: is \: \alpha < 90 \: so \: \\ \\ p = + 1$