Let f (θ) = 1 1 + (tan θ) ^ 2021 , then the value of ∑ _ θ = 1 ° ^ 89 ° f (θ) & is equal to

Correct answer: C. (89)/(2) f(θ) = (cos θ) ^ 2021 (sin θ) ^ 2021 + (cos θ) ^ 2021 f(90 ° - θ) = ((sin θ)2021)/((cos θ)2021 + (sin θ)2021) ⇒ f(θ) + f(90 ° - θ) = 1 and f(45 °) = (1)/(2) Hence, ∑ _ θ = 1 ^ o ^ 89 ^ o f(θ) = f( 1 ^ o ) + f( 2 ^ o ) + + f( 44 ^ o ) + f ( 46 ^ o ) + +f ( 88 ^ o ) +f( 89 ^ o ) +f( 45 ^ o ) = (f ( 1 ^ o ) ++f ( 89 ^ o )) + lleft(f( 2 ^ o ) + +f( 88 ^ o )) + + (f ( 44 ^ o ) + f ( 46 ^ o )) ++f( 45 ^ o ) = 1 ×44 + (1)/(2) = (89)/(2)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

NTA Abhyas 2020 · Mathematics · Trigonometric Ratios & Identities

Let f (θ) = 1 1 + (tan θ) ^ 2021 , then the value of ∑ _ θ = 1 ° ^...

Let

f

\left(\theta\right) = \dfrac{1}{1 + {

\left(tan \theta\right)}^{2021}},

then the value of

\mathop{∑}

_{\theta =

1 °}^{89 °}f \left(\theta\right)

& is equal to

Options

A $45$
B $44$
C $\dfrac{89}{2}$
D $\dfrac{91}{2}$

Correct answer

C. $\dfrac{89}{2}$

Step-by-step solution

f\left(\theta\right) = \dfrac{{\left(cos \theta\right)}^{2021}}{{\left(sin \theta\right)}^{2021} + {\left(cos \theta\right)}^{2021}}

f\left(90 ° - \theta\right) = \dfrac{\left(sin \theta\right)2021}{\left(cos \theta\right)2021 + \left(sin \theta\right)2021}

⇒ f\left(\theta\right) + f\left(90 ° - \theta\right) = 1

and

f\left(45 °\right) = \dfrac{1}{2}

Hence,

\mathop{∑}_{\theta = {1}^{o}}^{{89}^{o}} f\left(\theta\right) = f\left({1}^{o}\right) + f\left({2}^{o}\right) + \ldots + f\left({44}^{o}\right) + f \left({46}^{o}\right) + \ldots +f \left({88}^{o}\right) +f\left({89}^{o}\right) +f\left({45}^{o}\right)<br />=\lleft(f

\left({1}^{o}\right) ++f

\left({89}^{o}\right)) + \l

lleft(f\left({2}^{o}\right)

+

+f\left({88}^{o}\right)) + \ldots + \left(f \left({44}^{o}\right) +

f \left({46}^{o}\right)\right) ++f\left({45}^{o}\right)<

br />

= 1 \times44 + \dfrac{1}{2} = \dfrac{89}{2}

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