Explanation
Using,
\tan x=\cot \left( 90^{\circ}-x \right) and \tan x.\cot x=1
We get ,
\tan 5.\tan 10.\tan 15...\tan 80.\tan 85\\ =\tan 5.\tan 10.\tan 15...\tan 45. \cot(90-35)...\cot \left( 90-10 \right) .\cot \left( 90-5 \right) \\ =\tan 5.\cot 5.\tan 10.\cot 10...\tan 45\\ =1
In \triangle ABC, BC = 15, \sin B = \dfrac {4}{5} Now, \sin B = \dfrac {P}{H} \dfrac {4}{5} = \dfrac {AC}{AB} Then, let AC = 4x and AB = 5x Now, AB^2 = AC^2 + BC^2 (5x)^2 = (4x)^2 + 15^2 25x^2 = 16x^2 + 225 9x^2 = 225 x = 5 Hence, AC= 20 and AB = 25 \tan \angle ADC = 1 \dfrac {AC}{CD} = 1 AC = CD = 20 Now, using Pythagoras theorem, AD^2 = AC^2 + CD^2 AD^2 = 20^2 + 20^2 AD = 20 \sqrt{2}
In \triangle ABC, BC = 15, \sin B = \dfrac {4}{5} Now, \sin B = \dfrac {P}{H} \dfrac {4}{5} = \dfrac {AC}{AB} Then, let AC = 4x and AB = 5x Now, AB^2 = AC^2 + BC^2 (5x)^2 = (4x)^2 + 15^2 25x^2 = 16x^2 + 225 9x^2 = 225 x = 5 Hence, AC= 20 and AB = 25
\sin x+{ \sin }^{ 2 }x+{ \sin }^{ 3 }x=1\Rightarrow \sin x+{ \sin }^{ 3 }x=1-{ \sin }^{ 2 }x\\ \Rightarrow \sin x\left( 1+\sin ^{ 2 }x \right) ={ \cos }^{ 2 }x\Rightarrow \sin x\left( 2-{ \cos }^{ 2 }x \right) ={ \cos }^{ 2 }x\\ \Rightarrow { \sin }^{ 2 }x{ \left( 2-{ \cos }^{ 2 }x \right) }^{ 2 }={ \cos }^{ 4 }x\\ \Rightarrow \left( 1-{ \cos }^{ 2 }x \right) \left( 4+{ \cos }^{ 4 }x-4{ \cos }^{ 2 }x \right) ={ \cos }^{ 4 }x\\ \Rightarrow 4+{ \cos }^{ 4 }x-4{ \cos }^{ 2 }x-4{ \cos }^{ 2 }x-{ \cos }^{ 6 }x+4{ \cos }^{ 4 }x={ \cos }^{ 4 }x\\ \Rightarrow { \cos }^{ 6 }x-4{ \cos }^{ 4 }x+8{ \cos }^{ 2 }x=4
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