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).

Answer 2

Let us solve:

$\displaystyle{\frac{1}{1+tan^3\theta} +\frac{1}{1+cot^3\theta} }\\\\\displaystyle{= \frac{1}{1+tan^3\theta}+\frac{1}{1+\frac{1}{tan^3\theta} } }\\\\\displaystyle{= \frac{1}{1+tan^3\theta}+\frac{1}{\frac{tan^3\theta+1}{tan^3\theta} } }\\\\\\\displaystyle{= \frac{1}{1+tan^3\theta}+\frac{tan^3\theta}{1+tan^3\theta}}\\\\\displaystyle{= \frac{1+tan^3\theta}{1+tan^3\theta} }\\\\\bold{=1}$

$\underline{\textsf{On solving the identity, we get their sum as 1.}\\\\\bold{i.e.\ \frac{1}{1+tan^3\theta}+\frac{1}{1+cot^3\theta}=1 . }}$

$\mathbf{We\ know, tan(90\°-\theta)=cot\theta}\\\\\mathbf{And,\ cot(90\°-\theta)=tan\theta}$

$\bold{So,}\\\bullet\bold{tan80\° = {tan(90\°-10\°)} = cot10\°}\\\bullet\bold{tan70\° = {tan(90\°-20\°)} = cot20\°}\\\bullet\bold{tan60\° = tan(90\°-30\°) = cot30\°}\\\bullet\bold{tan50\° = tan(90\°-40\°) = cot40\°}$

$\underline{\textsf{The question given is:}}\\\\\displaystyle{\frac{1}{1+tan^3 10\°}+\frac{1}{1+tan^3 20\°} +\frac{1}{1+tan^3 30\°} +\frac{1}{1+tan^3 40\°}+\frac{1}{1+tan^3 50\°} +\frac{1}{1+tan^3 60\°} }\\\\\displaystyle{+\frac{1}{1+tan^3 70\°}+\frac{1}{1+tan^3 80\°} =? }$

$\textsf{\underline{Solving LHS:}}$

$\displaystyle{\frac{1}{1+tan^3 10\°}+\frac{1}{1+tan^3 20\°} +\frac{1}{1+tan^3 30\°} +\frac{1}{1+tan^3 40\°}+\frac{1}{1+cot^3 40\°} +\frac{1}{1+cot^3 30\°} }\\\\\displaystyle{+\frac{1}{1+cot^3 20\°}+\frac{1}{1+cot^3 10\°} }$$\\\textsf{From the identity we proved in the beginning, we get next:}\\\\\displaystyle{=1+1+1+1}\\\\\displaystyle{\boxed{=4}}$

So, the value we get is 4.