Question

If sin$^{2}$θ+2cos$^{2}$θ+3sin$^{2}$θ+4cos$^{2}$θ+......+ 40 terms = 405 where θ is acute then find the value of tan θ. ​

Answer 1

Solution:-

Given:-

$\rm \implies \: \sin {}^{2} \theta + 2 \cos^{2} \theta + 3 \sin^{2} \theta + 4 \cos ^{2} + ......... + 40 \: \: terms = 405$

Now separate Sin²θ and Cos²θ , We get

$\rm \implies \: \sin^{2} \theta + 3 \sin ^{2}\theta + 5 \sin^{2} \theta + ........... + 39 \sin^{2} \theta \: \: \: \: \: \: \: \: ......(i)eq$

Total Number of term ( n ) = 20

Now Take:-

$\rm \implies \: \cos^{2} \theta + 2 \cos^{2} \theta + 4 \cos^{2} \theta + ....... + 40 \cos^{2}\theta \: \: \: \: \: \: \: .........(ii)eq$

Total Number of term ( n )= 20

Now we can write both equation as

$\rm \implies \sin^{2}\theta(1 + 3 + 5 + ....... + 39) \: \: \: \: \: \: \: \: \: ....(i)eq$

$\rm \implies \cos^{2} \theta(2 + 4 + 6 + .......... + 40) \: \: \: \: \: \: \: ....(ii)eq$

Now find Sum of the number of both equation

Formula:-

$\rm \to \boxed{ \rm \: S_{n} = \dfrac{n}{2} \{2a + (n - 1)d \} = \dfrac{n}{2} (a + l)}$

Now Take

$\rm \implies \sin^{2}\theta(1 + 3 + 5 + ....... + 39) \: \: \: \: \: \: \: \: \: ....(i)eq$

$\rm \implies \: \sin ^{2} \theta \bigg( \dfrac{20}{2} \times \{1 + 39 \} \bigg)$

$\rm \implies \: \sin^{2} \theta(10 \times 40)$

$\rm \implies \: 400 \sin^{2} \theta$

Now Take

$\rm \implies \cos^{2} \theta(2 + 4 + 6 + .......... + 40) \: \: \: \: \: \: \: ....(ii)eq$

$\rm \implies \cos^{2} \theta \bigg( \dfrac{20}{2} \times \{2 + 40 \} \bigg)$

$\rm \implies \: \cos^{2} \theta (10 \times 42)$

$\rm \implies420 \cos^{2}\theta$

Its given Sum of term is 405

We can write as

$\rm \implies \: 400 \sin^{2}\theta + 420 \cos^{2}\theta = {405}$

Now simplify

$\rm \implies \: 400 \sin {}^{2} + 400 \cos ^{2} \theta + 20 \cos^{2} \theta = 405$

$\rm \implies \: 400( \sin^{2} \theta + \cos^{2}\theta) + 20 \cos {}^{2} \theta = 405$

$\rm \implies \: 400 + 20 \cos^{2} \theta = 405$

$\rm \implies \: 20 \cos {}^{2} \theta = 405 - 400$

$\rm \implies \cos^{2} \theta = \dfrac{5}{20}$

$\rm \implies \cos^{2} \theta = \dfrac{1}{4}$

$\rm \implies \cos \theta = \sqrt{ \dfrac{1}{4} }$

$\rm \implies \cos^{} \theta = \pm \dfrac{1}{2}$

It give angle will be acute so we take

$\rm \implies \cos^{} \theta = \dfrac{1}{2}$

$\rm \implies \theta = \cos^{ - 1} \bigg( \dfrac{1}{2} \bigg)$

$\rm \implies \: \theta = \dfrac{\pi}{3}$

We have to find value of Tanθ

$\rm \implies \tan\dfrac{\pi}{3} = \sqrt{3}$

Answer

$\rm \implies \tan\dfrac{\pi}{3} = \sqrt{3}$

Answer 2

sin2θ+cos2θ=1

sin2θ + b2cos2θ + 3sin2θ + 4cos2θ+....+40 terms = 405

⇒(sin2θ + 3sin2θ +... + 20bterms)+(2cos2θ+4cos2θ+...+20terms) = 405

⇒(1+3+5+...+20terms)+20cos2θ=405⇒202+20cos2θ=405⇒20cos2θ=5⇒cos2θ=41∴tanθ=3