Anyone plz answer this question ??
Answers
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
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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°
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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).
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Proving the identities-
↪{1/(1+tan³∅)}+{1/(1+cot³∅)}
↪{1/(1+tan³∅)}+[1/{(tan³∅)}]
↪{1/(1+tan³∅)}+[1+/{(tan³∅)}]
↪{1/(1+tan³∅)}+[1+/{tan³∅+1)/(tan³∅)}]
↪{1/(1+tan³∅)}+[tan³∅/(1+tan³∅)]
↪(1+tan³∅)/(1+tan³∅)
↪1
We can conclude that,
↪{1/(1+tan³∅)}+{1/(1+cot³∅)}=1
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1/(1+tan³10°)+(1+tan³20°)+1/(1+tan30°)_____1/(1+tan³80°)
As we know,tan(90-∅)=cot∅
Then,
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➡tan80° = {tan90°-10°} = cot10°
➡tan70° = {tan90°-20°} = cot20°
➡tan60° = {tan90°-30°} = cot30°
➡tan50° = {tan90°-40°} = cot40°
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By putting the given values here-
➡1/(tan³10°)+1/(1+tan³20°)+1/(1+tan³30°)+1/(1+tan³40°)+1/(1cot³30°)+1/(1+cot³20°)+1/(1+cot³10°)
➡ [1/(1+tan³10°)+1/(1+cot³10°)+1/(1+tan³20°)+1/(1+cot³20°)+1/(1+tan³30°)+1/(1cot³30°)+(1+tan³40°)+(1+cot³40°)]
Now,by using our above identity,,we get-
➡ 1+1+1+1