Question

Anyone plz answer this question ??​

Answer 1

Solution :-

First Lets Prove a Identity :-

→ {1/(1 + tan³θ)} + {1/(1 + cot³θ)}

Putting cotθ = (1/tanθ) we get,

→ {1/(1 + tan³θ)} + [ 1 + /{ 1 + (1/tan³θ)} ]

Taking LCM of Second Part ,

→ {1/(1 + tan³θ)} + [ 1 + / { (tan³θ + 1) / tan³θ } ]

→ {1/(1 + tan³θ)} + {tan³θ / (1 + tan³θ)}

Taking LCM on Both Now,

→ (1 + tan³θ) / (1 + tan³θ)

→ 1.

So, we can Conclude That,

☛ {1/(1 + tan³θ)} + {1/(1 + cot³θ)} = 1

_______________________

Question :-

➺ 1/(1+tan³10°) + 1/(1+tan³20°) + 1/(1+tan30°) ______________ 1/(1+tan³80°)

we know That,

☞ tan(90-θ) = cotθ

So,

☞ tan80° = tan(90°-10°) = cot10°

☞ tan70° = tan(90°-20°) = cot20°

☞ tan60° = tan(90°-30°) = cot30°

☞ tan50° = tan(90-40°) = cot40°

_________________________

Putting These values , we can say That, we have To Find Value of :-

➼ 1/(1+tan³10°) + 1/(1+tan³20°) + 1/(1+tan³30°) + 1/(1+tan³40°) + 1/(1+cot³40°) + 1/(1+cot³30°) + 1/(1+cot³20°) + 1/(1+cot³10°)

Re - arranging Them now, we get,

➼ [ 1/(1+tan³10°) + 1/(1+cot³10°) ] + [ 1/(1+tan³20°) + 1/(1+cot³20°) ] + [ 1/(1+tan³30°) + 1/(1+cot³30°) ] + [ 1/(1+tan³40°) + 1/(1+cot³40°) ]

Using Our Proving Identity Now , we get,

☛ 1 + 1 + 1 + 1

☛ 4 (Ans).

[ Nice Question . ]

Answer 2

Given,

$\displaystyle\longrightarrow\sf{S=\dfrac{1}{1+\tan^310^o}+\dfrac{1}{1+\tan^320^o}+\dfrac{1}{1+\tan^330^o}+\,\dots\,+\dfrac{1}{1+\tan^380^o}}$

First consider first and last terms in the sum together.

$\displaystyle\longrightarrow\sf{\dfrac{1}{1+\tan^310^o}+\dfrac{1}{1+\tan^380^o}=\dfrac{1+\tan^310^o+1+\tan^380^o}{(1+\tan^310^o)(1+\tan^380^o)}}$

$\displaystyle\longrightarrow\sf{\dfrac{1}{1+\tan^310^o}+\dfrac{1}{1+\tan^380^o}=\dfrac{\tan^310^o+\tan^380^o+2}{1+\tan^310^o+\tan^380^o+\tan^310^o\tan^380^o}}$

But we see that,

• $\sf{\tan^380^o=\cot^310^o}$

because we know that $\sf{\tan(90^o-\theta)=\cot\theta.}$ Then,

$\displaystyle\longrightarrow\sf{\dfrac{1}{1+\tan^310^o}+\dfrac{1}{1+\tan^380^o}=\dfrac{\tan^310^o+\cot^310^o+2}{1+\tan^310^o+\cot^310^o+\tan^310^o\cot^310^o}}$

Since $\tan\theta$ and $\cot\theta$ are reciprocals to each other,

$\displaystyle\longrightarrow\sf{\dfrac{1}{1+\tan^310^o}+\dfrac{1}{1+\tan^380^o}=\dfrac{\tan^310^o+\cot^310^o+2}{1+\tan^310^o+\cot^310^o+1}}$

$\displaystyle\longrightarrow\sf{\dfrac{1}{1+\tan^310^o}+\dfrac{1}{1+\tan^380^o}=\dfrac{\tan^310^o+\cot^310^o+2}{\tan^310^o+\cot^310^o+2}}$

$\displaystyle\longrightarrow\sf{\dfrac{1}{1+\tan^310^o}+\dfrac{1}{1+\tan^380^o}=1}$

Similarly we get,

$\displaystyle\longrightarrow\sf{\dfrac{1}{1+\tan^320^o}+\dfrac{1}{1+\tan^370^o}=1}$

$\displaystyle\longrightarrow\sf{\dfrac{1}{1+\tan^330^o}+\dfrac{1}{1+\tan^360^o}=1}$

$\displaystyle\longrightarrow\sf{\dfrac{1}{1+\tan^340^o}+\dfrac{1}{1+\tan^350^o}=1}$

Hence,

$\displaystyle\longrightarrow\sf{S=\dfrac{1}{1+\tan^310^o}+\dfrac{1}{1+\tan^320^o}+\dfrac{1}{1+\tan^330^o}+\,\dots\,+\dfrac{1}{1+\tan^380^o}}$

$\displaystyle\begin{aligned}\longrightarrow\ \ &\mathsf{S=\left(\dfrac{1}{1+\tan^310^o}+\dfrac{1}{1+\tan^380^o}\right)}\\&\ \ +\mathsf{\left(\dfrac{1}{1+\tan^320^o}+\dfrac{1}{1+\tan^370^o}\right)}\\&\ \ +\sf{\left(\dfrac{1}{1+\tan^330^o}+\dfrac{1}{1+\tan^360^o}\right)}\\&\ \ +\sf{\left(\dfrac{1}{1+\tan^340^o}+\dfrac{1}{1+\tan^350^o}\right)}\end{aligned}$

$\displaystyle\longrightarrow\sf{S=1+1+1+1}$

$\displaystyle\longrightarrow\sf{\underline{\underline{S=4}}}$

Hence 4 is the answer.