Question

without using trigonometry tables, evaluate

(tan20°/cosec70°)^2+(cot20°/sec70°)^2+2 tan15°tan37°tan53°tan60°tan75°

​

Answer 1

Step-by-step explanation:

answer is 3

if u want process I will send

Answer 2

$\bf{\gray{\underbrace{\blue{TO\:FIND:-}}}}$

$\red\bullet\:\bf{\Big(\dfrac{\tan{20^{\degree}}}{\cosec{70^{\degree}}}\Big)^2\:+\:\Big(\dfrac{\cot{20^{\degree}}}{\sec{70^{\degree}}}\Big)^2\:+\:2\:.\:\tan{15^{\degree}}\:.\:\tan{37^{\degree}}\:.\:\tan{53^{\degree}}\:.\:\tan{60^{\degree}}\:.\:\tan{75^{\degree}}\:}$

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$\bf{\gray{\underbrace{\blue{Trigonometric\:FORMULA:-}}}}$

$1.\:\bf\red{\cosec{(90^{\degree}\:-\:\theta)}\:=\:\sec{\theta}\:}$

$2.\:\bf\purple{\sec{(90^{\degree}\:-\:\theta)}\:=\:\cosec{\theta}\:}$

$3.\:\bf\orange{\tan{(90^{\degree}\:-\:\theta)}\:=\:\cot{\theta}\:}$

$4.\:\bf\green{\tan{\theta}\:.\:\cot{\theta}\:=\:1\:}$

$5.\:\bf\blue{\cosec{\theta}\:=\:{\dfrac{1}{\sin{\theta}}}\:}$

$6.\:\bf{\sec{\theta}\:=\:{\dfrac{1}{\cos{\theta}}}\:}$

$7.\:\bf\pink{\tan{\theta}\:=\:{\dfrac{\sin{\theta}}{\cos{\theta}}}\:}$

$8.\:\bf\gray{\cot{\theta}\:=\:{\dfrac{\cos{\theta}}{\sin{\theta}}}\:}$

$9.\:\bf\red{\sin^2{\theta}\:+\:\cos^2{\theta}\:=\:1\:}$

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$\bf{\gray{\underbrace{\blue{SOLUTION:-}}}}$

$\rm{:\implies\:\Big(\dfrac{\tan{20^{\degree}}}{\cosec{70^{\degree}}}\Big)^2\:+\:\Big(\dfrac{\cot{20^{\degree}}}{\sec{70^{\degree}}}\Big)^2\:+\:2\:.\:\tan{15^{\degree}}\:.\:\tan{37^{\degree}}\:.\:\tan{53^{\degree}}\:.\:\tan{60^{\degree}}\:.\:\tan{75^{\degree}}\:}$

$\rm{:\implies\:\Big\{\dfrac{\tan{20^{\degree}}}{\cosec{(90^{\degree}\:-\:20^{\degree})}}\Big\}^2\:+\:\Big\{\dfrac{\cot{20^{\degree}}}{\sec{(90^{\degree}\:-\:20^{\degree})}}\Big\}^2\:+\:2\:.\:\tan{15^{\degree}}\:.\:\tan{37^{\degree}}\:.\:\tan{(90^{\degree}\:-\:37^{\degree})}\:.\:\tan{60^{\degree}}\:.\:\tan{(90^{\degree}\:-\:15^{\degree})}\:}$

$\rm{:\implies\:\Big(\dfrac{\tan{20^{\degree}}}{\sec{20^{\degree}}}\Big)^2\:+\:\Big(\dfrac{\cot{20^{\degree}}}{\cosec{20^{\degree}}}\Big)^2\:+\:2\:.\:\cancel{\tan{15^{\degree}}}\:.\:\cancel{\tan{37^{\degree}}}\:.\:\cancel{\cot{37^{\degree}}}\:.\:\tan{60^{\degree}}\:.\:\cancel{\cot{15^{\degree}}}\:}$

$\rm{:\implies\:\Big\{\dfrac{\sin{20^{\degree}}/\cancel{\cos{20^{\degree}}}}{1/\cancel{\cos{20^{\degree}}}}\Big\}^2\:+\:\Big\{\dfrac{\cos{20^{\degree}}/\cancel{\sin{20^{\degree}}}}{1/\cancel{\sin{20^{\degree}}}}\Big\}^2\:+\:2\:.\:\tan{60^{\degree}}\:}$

$\rm{:\implies\:\sin^220^{\degree}\:+\:\cos^220^{\degree}\:+\:2\:.\:\tan{60^{\degree}}\:}$

$\rm{:\implies\:1\:+\:2\times{\sqrt{3}}\:}$

$\bf\green{:\implies\:1\:+\:2\sqrt{3}\:}$

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$\red\therefore\:\bf{Required\:Answer\:\longrightarrow\:\purple{1\:+\:2\sqrt{3}}\:}$