Trigonometry problem #4

Geometry Level 2

True or false:

tan ⁡ β sec ⁡ β + 1 − sec ⁡ β − 1 tan ⁡ β = 1 \large \frac{\tan \beta}{\sec \beta +1} - \frac{\sec \beta -1}{\tan \beta} = 1

False True

Munem Shahriar by Munem Shahriar , 18, Bangladesh

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3 solutions

Munem Shahriar
Feb 18, 2018

tan ⁡ β sec ⁡ β + 1 − sec ⁡ β − 1 tan ⁡ β = tan ⁡ 2 β − ( sec ⁡ β + 1 ) ( sec ⁡ β − 1 ) ( sec ⁡ β + 1 ) ( tan ⁡ β ) = tan ⁡ 2 β − ( sec ⁡ 2 β − 1 ) ( tan ⁡ β ) ( sec ⁡ β + 1 ) = tan ⁡ 2 β − tan ⁡ 2 β ( tan ⁡ β ) ( sec ⁡ β + 1 ) = 0 ( tan ⁡ β ) ( sec ⁡ β + 1 ) = 0 \begin{aligned} \dfrac{\tan \beta}{\sec \beta +1} - \dfrac{\sec \beta -1}{\tan \beta} &= \dfrac{\tan^2 \beta - (\sec \beta + 1) (\sec \beta - 1)}{(\sec \beta + 1)(\tan \beta)} \\& = \dfrac{\tan ^2 \beta - (\sec^2 \beta - 1)}{(\tan \beta)(\sec \beta +1)} \\& = \dfrac{\tan^2 \beta - \tan^2 \beta}{(\tan \beta)(\sec \beta +1)} \\& = \dfrac 0{(\tan \beta)(\sec \beta + 1)} \\& = 0 \\ \end{aligned}

Hence 0 ≠ 1 0 \ne 1

Note: sec ⁡ 2 β − 1 = tan ⁡ 2 β \sec^2 \beta - 1 = \tan^2 \beta

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Tom Engelsman
Feb 17, 2018

If we apply a common denominator, we obtain:

t a n ( β ) s e c ( β ) + 1 − s e c ( β ) − 1 t a n ( β ) = t a n 2 ( β ) − ( s e c 2 ( β ) − 1 ) t a n ( β ) ( s e c ( β ) + 1 ) \frac{tan(\beta)}{sec(\beta) + 1} - \frac{sec(\beta) -1}{tan(\beta)} = \frac{tan^{2}(\beta) - (sec^{2}(\beta) - 1)}{tan(\beta)(sec(\beta) + 1)} ;

but s e c 2 ( β ) − t a n 2 ( β ) = 1 sec^{2}(\beta) - tan^{2}(\beta) = 1 , which yields zero in the numerator. Hence, the answer is 0 . \boxed{0}.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

tan ⁡ β sec ⁡ β + 1 − sec ⁡ β − 1 tan ⁡ β \frac {\tan\beta}{\sec\beta +1} - \frac {\sec\beta - 1}{\tan\beta}

= tan ⁡ 2 β − ( sec ⁡ 2 β − 1 ) tan ⁡ β ( sec ⁡ β + 1 ) = \frac {\tan^2 \beta - (\sec^2 \beta -1)}{\tan\beta (\sec\beta +1)}

= tan ⁡ 2 β − sec ⁡ 2 β + 1 tan ⁡ β ( sec ⁡ β + 1 ) = \frac {\tan^2\beta - \sec^2\beta +1}{\tan\beta (\sec\beta +1)}

= − 1 + 1 tan ⁡ β ( sec ⁡ β + 1 ) = \frac {-1+1}{\tan\beta (\sec\beta +1) }

= 0. =0.

But, 0 ≠ 1. 0 \neq 1. Therefore, the answer is F a l s e . \boxed {False.}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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