Sec'o'tan!

Geometry Level 3

sec ⁡ θ + tan ⁡ θ = 1 \large \sec \theta + \tan \theta = 1

Find the value of sec ⁡ θ \sec \theta .

Give your answer ot 3 decimal places.


The answer is 1.0000000000000000000000000000000.

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Viki Zeta by Viki Zeta , 19, India

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1 solution

Viki Zeta
Dec 31, 2016

There are many ways to solve this, mainly using identities. I've did this in a different way prettty easy using linear equation rules. sec ⁡ ( θ ) + tan ⁡ ( θ ) = 1 sec ⁡ ( θ ) + tan ⁡ ( θ ) = sec ⁡ 2 ( θ ) − tan ⁡ 2 ( θ ) sec ⁡ ( θ ) [ 1 − sec ⁡ ( θ ) ] + tan ⁡ ( θ ) [ 1 + tan ⁡ ( θ ) ] = 0 sec ⁡ ( θ ) [ 1 − sec ⁡ ( θ ) ] + tan ⁡ ( θ ) [ 1 + tan ⁡ ( θ ) ] = sec ⁡ ( θ ) × 0 + tan ⁡ ( θ ) × 1 ⟹ 1 − sec ⁡ ( θ ) = 0 ⟹ sec ⁡ ( θ ) = 1 \sec(\theta) + \tan(\theta) = 1 \\ \sec(\theta) + \tan(\theta) = \sec^2(\theta) - \tan^2(\theta) \\ \sec(\theta) [ 1 - \sec(\theta)] + \tan(\theta) [ 1 + \tan(\theta) ] = 0\\ \sec(\theta) [ 1 - \sec(\theta)] + \tan(\theta) [ 1 + \tan(\theta) ] = \sec(\theta) \times 0 + \tan(\theta) \times 1 \\ \implies 1 - \sec(\theta) = 0 \\ \implies \sec(\theta) = 1

Here I've assumed θ \theta to be any angle, but I've not included 0 in the coefficient of tan ⁡ \tan on RHS because tan ⁡ \tan holds a value of 0 at tan ⁡ 0 \tan 0 , whereas it's impossible for sec ⁡ \sec to hold any angle 0 0 .

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But I've some problem i.e. secβ+tanβ=1 then, secβ-tanβ=1/3

So, by simply adding both we get 2secβ=4/3 so, secβ=2/3=0.6666667

Adarsh Mahor - 4 years, 5 months ago

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If sec ⁡ x + tan ⁡ x = 1 \sec x + \tan x = 1 how would sec ⁡ x − tan ⁡ x = 1 3 \sec x - \tan x = \dfrac{1}{3} ?

sec ⁡ 2 x − tan ⁡ 2 x = 1 ( sec ⁡ x + tan ⁡ x ) ( sec ⁡ x − tan ⁡ x ) = 1 sec ⁡ x − tan ⁡ x = 1 \sec^2x - \tan^2x=1 \\ (\sec x + \tan x)(\sec x - \tan x) = 1 \\ \sec x - \tan x = 1

Viki Zeta - 4 years, 5 months ago

It is impossible what you are saying Adarsh , as s e c 2 θ − t a n 2 θ sec^2\theta - tan^2\theta = 1

Now s e c θ + t a n θ sec\theta + tan\theta = 1

So s e c θ − t a n θ sec\theta - tan \theta = 1 is obvbious

Md Zuhair - 4 years, 5 months ago

Lovely desinged answer :P

Md Zuhair - 4 years, 5 months ago

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