Question

plz help me plz and correctly!!!! only 3 questions plz!!!​ 100 points

Answer 1

Answer:

Step-by-step explanation:

6)b) Sin A - x Cos A = 0

Sin A = x Cos A

$\dfrac{Sin \ A}{Cos \ A} = x$

$\dfrac{1}{x}=\dfrac{1}{\dfrac{Sin \ A }{Cos \ A}}=1*\dfrac{Cos \ A}{Sin \ A}=\dfrac{Cos \ A}{Sin \ A}$

$RHS = x +\dfrac{1}{x}$

         $= \dfrac{Sin \ A}{Cos \ A}+\dfrac{Cos \ A}{Sin \ A}\\\\\\=\dfrac{Sin \ A*Sin \ A}{Cos \ A*Sin \ A}+\dfrac{Cos \ A * Cos \ A}{Sin \ A* Cos \ A}\\\\\\= \dfrac{Sin^{2} \ A}{Cos \ A*Sin \ A}+\dfrac{Cos^{2} \ A}{Cos \ A* Sin \ A}\\\\= \dfrac{Sin^{2} \ A+Cos^{2} \ A}{Cos \ A*Sin \ A}\\\\=\dfrac{1}{Cos \A*Sin \ A}\\\\= \dfrac{1}{Cos \ A}*\dfrac{1}{Sin \ A}\\\\=Sec \ A * Cosec \ A$

          = LHS

7a) 1 - Sin A = 1/2

     $1 = \dfrac{1}{2}+ Sin \ A\\\\1-\dfrac{1}{2}=Sin \ A\\\\\dfrac{1}{2}=Sin \ A$

⇒ A = 30°

Sec² A = Sec² 30° = $(\frac{2}{\sqrt{3}})^{2}=\frac{2^{2}}{(\sqrt{3)^{2}}}=\frac{4}{3}$

$tan^{2} \ A = tan^{2} 30 = (\dfrac{1}{\sqrt{3}})^{2} = \dfrac{1}{3}$

LHS = 6 Sec² A + 9 tan² A

       $= 6*\dfrac{4}{3}+9*\dfrac{1}{3}\\\\\\=2*4 + 3*1\\\\\\= 8 + 3\\\\= 11 = RHS$