Question

Cos 70 degree divided by sin 20 degree Plus cos 55 degree * cosec 35 degree divided by tan 5 degree into tan 25 degree into tan 65 degree into tan 85 degree is equal to 2

Answer 1

Answer:

To prove:  

 $\frac{cos70^{\circ}}{sin20^{\circ}}+\frac{cos55^{\circ}\times cosec35^{\circ}}{tan5^{\circ}\times tan25^{\circ}\times tan65^{\circ}\times tan85^{\circ}}=2$

Consider,

$LHS=\frac{cos70^{\circ}}{sin20^{\circ}}+\frac{cos55^{\circ}\times cosec35^{\circ}}{tan5^{\circ}\times tan25^{\circ}\times tan65^{\circ}\times tan85^{\circ}}$

$=\frac{cos70^{\circ}}{sin(90^{\circ}-70^{\circ})}+\frac{cos55^{\circ}\times\frac{1}{sin35^{\circ}}}{tan5^{\circ}\times tan25^{\circ}\times tan(90^{\circ}-25^{\circ})\times tan(90^{\circ}-5^{\circ})}$

$=\frac{cos70^{\circ}}{sin(90^{\circ}-70^{\circ})}+\frac{cos55^{\circ}\times\frac{1}{sin(90^{\circ}-55^{\circ})}}{tan5^{\circ}\times tan25^{\circ}\times tan(90^{\circ}-25^{\circ})\times tan(90^{\circ}-5^{\circ})}$

Now use trigonometry complementary angle,

       sin ( 90° - x ) = cos x  and  tan ( 90° - x ) = cot x , we get

$=\frac{cos70^{\circ}}{cos70^{\circ}}+\frac{cos55^{\circ}\times\frac{1}{cos55^{\circ}}}{tan5^{\circ}\times tan25^{\circ}\times cot25^{\circ}\times cot5^{\circ}}$

$=1+\frac{1}{tan5^{\circ}\times tan25^{\circ}\times cot25^{\circ}\times cot5^{\circ}}$

Now use, result, $tan\theta\times cot\theta=1$ we get

$=1+\frac{1}{1}$

= 1 + 1

= 2

⇒ LHS = RHS

Hence Proved

Answer 2

Answer:

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