Question

the value of [ sin π/9] [4+sec π/9]​

Answer 1

Step-by-step explanation:

hopefully it would help you..

Jai Hind

Answer 2

Answer:

Concept:

Simplifying and finding the value of given statement.

Step-by-step explanation:

Given:

$[ sin \frac{\pi}{9}] [4+sec \frac{\pi}{9}]$

To find:

Value of  $[ sin \frac{\pi}{9}] [4+sec \frac{\pi}{9}]$

Solution:

• Value of $\pi=180^\circ$
• Given statement contains $\frac{\pi}{9}$

i.e. $\frac{\pi}{9} = \frac{180^\circ}{9}$

     $\frac{\pi}{9} = 20^\circ$

• $sin \frac{\pi}{9} = sin 20^\circ$    and  $sec\frac{\pi}{9} = sec20^\circ$
• Substituting these,

$[ sin \frac{\pi}{9}] [4+sec \frac{\pi}{9}] =sin 20^\circ(4+ sec20^\circ)$

        $=sin 20^\circ(4+ \frac{1}{cos20^\circ} )$

          $=sin 20^\circ(\frac{4cos20^\circ+1}{cos20^\circ} {} )$

          $=\frac{4sin 20^\circ\cos20^\circ+sin 20^\circ}{cos20^\circ}$

            $=\frac{2sin 40^\circ+sin 20^\circ}{cos20^\circ}$        (since , $2sin\theta cos\theta=sin2\theta$)

            $=\frac{2sin (60-20)^\circ+sin 20^\circ}{cos20^\circ}$

• sing the formula,

$sin(A-B)=sinAcosB-cosAsinB$

 $\frac{2sin (60-20)^\circ+sin 20^\circ}{cos20^\circ} =\frac{2(sin60^\circ cos20^\circ-cos60^\circ sin20)^\circ+sin 20^\circ}{cos20^\circ}$

                              $=\frac{2(sin60^\circ cos20^\circ)-2(cos60^\circ sin20)^\circ+sin 20^\circ}{cos20^\circ}$

                               $=\frac{2(sin(\frac{\sqrt{3} }{2}) cos20^\circ)-2(cos(\frac{1}{2} )sin20^\circ+sin 20^\circ}{cos20^\circ}$

                               $=\frac{2(\frac{\sqrt{3} }{2}) cos20^\circ}{cos20^\circ} -\frac{2(\frac{1}{2} )sin20^\circ}{cos20^\circ} +\frac{sin 20^\circ}{cos20^\circ}$

                                $=2(\frac{\sqrt{3} }{2}) -\frac{sin20^\circ}{cos20^\circ} +\frac{sin 20^\circ}{cos20^\circ}$

                                  $=\sqrt{3}$

Hence, value is $\sqrt{3}$            

#SPJ3