Five storey Trigonometry Tower

Geometry Level 2

sin ⁡ x cos ⁡ x tan ⁡ x cot ⁡ x sec ⁡ x csc ⁡ x \huge \frac{\hspace{2mm} \frac{\hspace{2mm} \frac{\hspace{2mm} \frac{\hspace{2mm} \frac{\sin x}{\cos x}\hspace{2mm} }{\tan x}\hspace{2mm} }{\cot x}\hspace{2mm} }{\sec x}\hspace{2mm} }{\csc x}

Find the value of the above fraction tower if 0 ∘ < x < 90 ∘ 0^\circ < x < {90}^\circ .

0 1 sin ⁡ 2 x \sin^2 x cos ⁡ 2 x \cos^2 x tan ⁡ x \tan x Undefined

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Nihar Mahajan by Nihar Mahajan , 20, India

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5 solutions

Chew-Seong Cheong
Jun 24, 2015

( ( ( ( sin ⁡ x cos ⁡ x ) tan ⁡ x ) cot ⁡ x ) sec ⁡ x ) csc ⁡ x = ( ( ( tan ⁡ x tan ⁡ x ) cot ⁡ x ) sec ⁡ x ) csc ⁡ x = ( ( 1 cot ⁡ x ) sec ⁡ x ) csc ⁡ x = ( tan ⁡ x sec ⁡ x ) csc ⁡ x = sin ⁡ x csc ⁡ x = sin ⁡ 2 x \begin{aligned} \frac{\left( \frac {\left( \frac{\left(\frac{\left( \frac{\sin{x}}{\cos{x}} \right)} {\tan{x}} \right)}{\cot{x}} \right)} {\sec{x}} \right) }{\csc{x}} & = \frac{\left( \frac {\left( \frac{\left(\frac{\tan{x}} {\tan{x}} \right)}{\cot{x}} \right)} {\sec{x}} \right) }{\csc{x}} = \frac{\left( \frac {\left( \frac{1}{\cot{x}} \right)} {\sec{x}} \right) }{\csc{x}} = \frac{\left( \frac {\tan{x}} {\sec{x}} \right) }{\csc{x}} \\ & = \frac {\sin{x}}{\csc{x}} = \boxed{\sin^2{x}} \end{aligned}

47 Helpful 0 Interesting 0 Brilliant 1 Confused

Moderator note:

Simple standard approach.

what exactky is wrong if i do these: (sinx/cosx ) div (tanx/cotx/ secx/cscx)=(sinx/cosx) div ( (tanx/cotx) div (secx/cscx)) = (sinx/ cosx) div (tanx ^2 * cscx/ secx) = (sinx/cosx ) * secx / (cscx * tanx ^2 ) = sinx ^2 / (cosx ^2 * tanx ^2) = 1

the position of equal to sign "=" vertically aligned to which of the horizontal division lines... is Important... which is not clearly mentioned in the problem.

.. :/ (meh)

Ananya Aaniya - 5 years, 11 months ago

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Different steps will have different results. The step should start from top to bottom as shown in my solution.

Chew-Seong Cheong - 5 years, 11 months ago

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What is the main reasoning behind your word "should"? how can one arbitrarily be sure of what is "should" and what "should not" when no clear specification in qs?

Ananya Aaniya - 5 years, 11 months ago

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@Ananya Aaniya – Sorry, for not using the right word. I mean "by convention". For example: 5 4 3 2 1 5^{4^{3^{2^{1}}}} , the calculation should, "oops", I mean "must" start from 2 1 2^1 , then 3 2 3^2 , 4 9 4^9 then 5 2 62144 5^262144 . Here, it should "oops", must start from sin ⁡ x / cos ⁡ x \sin{x}/\cos{x} , then tan ⁡ x / tan ⁡ x \tan{x}/\tan{x} , 1 / cot ⁡ x 1/ \cot{x} , tan ⁡ x / sec ⁡ x \tan{x}/\sec{x} ,.... I should be going out for lunch now.

Chew-Seong Cheong - 5 years, 11 months ago

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@Chew-Seong Cheong – @Chew-Seong Cheong Haha! your so funny :-) nice humor

Maygrens Macatangay - 4 years, 6 months ago

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@Maygrens Macatangay – I mean "you're"

Maygrens Macatangay - 4 years, 6 months ago
Kelly B
Jul 4, 2015

17 Helpful 0 Interesting 1 Brilliant 0 Confused

This definitely is smooth!

Aidan Zoldyk - 3 years, 6 months ago
Lew Sterling Jr
Jul 5, 2015

4 Helpful 1 Interesting 0 Brilliant 0 Confused
Chris Brown
Apr 20, 2020

Starting at the top, sin ⁡ ( x ) cos ⁡ ( x ) = tan ⁡ ( x ) \frac { \sin { \left( x \right) } }{ \cos { \left( x \right) } } =\tan { \left( x \right) } . That leaves us with tan ⁡ ( x ) tan ⁡ ( x ) = 1 \frac { \tan { \left( x \right) } }{ \tan { \left( x \right) } } =1 . This subsequently leaves us with 1 cot ⁡ ( x ) = tan ⁡ ( x ) \frac { 1 }{ \cot { \left( x \right) } } =\tan { \left( x \right) } . This then leaves us with tan ⁡ ( x ) sec ⁡ ( x ) \frac { \tan { \left( x \right) } }{ \sec { \left( x \right) } } which, if you do some exploring with the ratios of the sides, is equal to sin ⁡ ( x ) \sin { \left( x \right) } . This finally leaves us with sin ⁡ ( x ) csc ⁡ ( x ) = sin ⁡ 2 ( x ) \frac { \sin { \left( x \right) } }{ \csc { \left( x \right) } } =\sin ^{ 2 }{ \left( x \right) } .

0 Helpful 1 Interesting 0 Brilliant 0 Confused
Jack Rawlin
Feb 10, 2016

( sin ⁡ x cos ⁡ x ) tan ⁡ x cot ⁡ x sec ⁡ x csc ⁡ x , sin ⁡ x cos ⁡ x = tan ⁡ x \Large \frac{\frac{\frac{\frac{\left(\frac{\sin x}{\cos x}\right)}{\tan x}}{\cot x}}{\sec x}}{\csc x},~ \frac{\sin x}{\cos x} = \tan x

( tan ⁡ x tan ⁡ x ) cot ⁡ x sec ⁡ x csc ⁡ x \Large \frac{\frac{\frac{\left(\frac{\tan x}{\tan x}\right)}{\cot x}}{\sec x}}{\csc x}

( 1 cot ⁡ x ) sec ⁡ x csc ⁡ x , 1 cot ⁡ x = tan ⁡ x \Large \frac{\frac{\left(\frac{1}{\cot x}\right)}{\sec x}}{\csc x},~ \frac{1}{\cot x} = \tan x

( tan ⁡ x sec ⁡ x ) csc ⁡ x , tan ⁡ x sec ⁡ x = tan ⁡ x cos ⁡ x = sin ⁡ x \Large \frac{\left(\frac{\tan x}{\sec x}\right)}{\csc x},~ \frac{\tan x}{\sec x} = \tan x \cos x = \sin x

sin ⁡ x csc ⁡ x , sin ⁡ x sin ⁡ x = sin ⁡ 2 x \Large \frac{\sin x}{\csc x},~ \sin x \sin x = \sin^2 x

sin ⁡ 2 x \Large \boxed{\sin^2 x}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Anyone pls tell me the meaning of cscx in this question.? Is cscx=cosx÷sinx÷cosx?

Hitixa Raj - 4 years, 6 months ago

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csc ⁡ ( x ) \csc(x) is short for c o s e c ( x ) cosec(x) which is equal to 1 sin ⁡ ( x ) \frac{1}{\sin(x)} .

So

sin ⁡ ( x ) csc ⁡ ( x ) = sin ⁡ ( x ) 1 sin ⁡ ( x ) = sin ⁡ ( x ) ⋅ sin ⁡ ( x ) 1 = sin ⁡ 2 ( x ) \frac{\sin(x)}{\csc(x)} = \frac{\sin(x)}{\frac{1}{\sin(x)}} = \sin(x) \cdot \frac{\sin(x)}{1} = \sin^2(x)

Jack Rawlin - 4 years, 6 months ago

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