Question

answer the question​

Answer 1

Answer:

7

Step-by-step explanation:

Given that $4x^{2} -3x+\alpha =0$ has two roots as sinA and cosA

so sinA+cosA= 3/4  and  sinAcosA=$\frac{\alpha}{4}$

Given sinA+cosA+tanA+cotA+secA+cosecA=7

$\frac{3}{4} +\frac{sinA}{cosA} +\frac{cosA}{sinA} +\frac{1}{cosA} +\frac{1}{sinA} =7$

$\frac{3}{4} + \frac{sinA^{2} +cosA^{2} }{sinAcosA} +\frac{sinA+cosA}{sinAcosA} =7$

$\frac{1}{\frac{\alpha }{4} } + \frac{3/4}{\alpha/4 } =7$

3/alpha+4/alpha=25/4

7/alpha=25/4

alpha=28/25

hence value of $\frac{25\alpha }{4}$= 25*28/25*1/4=7