please answer this 3 question with (step by step)please please please

1.	please answer this 3 question with (step by step)please please please please please please please please please please please please please please please please
Answer» Answer: 1) GIVEN: tan 53° . tan 50° . tan 37° . tan 40° = 1 We know that tan and cot are complementary (i.e. sum is equal to 90°) $\tan(90 ^{ \circ} - \theta) = \cot \: \theta$ $\therefore \: \tan( {90}^{\circ} - {37}^{\circ} ) = \cot \: {37}^{\circ}$ and $\tan \: ( {90}^{\circ} - {50}^{\circ} ) = \cot \: {50}^{\circ}$ Therefore, LHS $\tan \: {37}^{\circ} \times \cot \: {37}^{\circ} \times\tan \: {50}^{\circ} \times \cot \: {50}^{\circ}$ We know that tan and cot are reciprocal of each other. $\tan \: {37}^{\circ} \times \frac{1}{\tan {37}^{\circ} } \times \tan {50}^{\circ} \times \frac{1}{\tan {50}^{\circ} }$ $1 \times 1$ $1$ Therefore, LHS = RHS 2) Given: Two points : (-15,0) and (6,0) Here, $x_1 = -15$ $y_1 = 0$ $x_2 =6$ $y_2 = 0$ $\tt \: Distance \: Formula = \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} }$ Substituting the VALUES , we get : $\tt \: distance = \sqrt{(6 - ( - 15))^{2} + {(0 - 0)}^{2} }$ $\tt \: distance \: = \sqrt{ {21}^{2} }$ $\tt\therefore \: distance = 21 \: units$ 3) Given: $\frac{4 {x}^{2} + 3 \sqrt{3}x + 7 }{7} = 0$ Cross MULTIPLYING, we get: $4 {x}^{2} + 3 \sqrt{3} x + 7 = 0$ Here, a = 4 b = 3√3 c = 7