ත්‍රිකෝණමිතිය ගාණක්

[poopoo]

මේක හදලා කවුරුහරි

 

[zid2021]

Write in terms sin cos and cancel out terms

 

[acryan]

COTA = 1/ Tan A dala balanna enawa bung

Okkoma Tan walata harawanna. mama me dan haduwa nikan mathakada balanna

 

[kinkon]

ත්‍රිකෝණ

 

[lasankandy]

Plan: 1. We know that cot⁡(A)=1tan⁡(A) and tan⁡(A)=1cot⁡(A). We can use these identities to simplify the given equation. 2. After simplifying, we should be able to prove the given identity.

Let's carry out the plan:

Step 1: Substitute cot⁡(A)=1tan⁡(A) and tan⁡(A)=1cot⁡(A) in the given equation:

1tan⁡(A)+tan⁡(B)1tan⁡(B)+tan⁡(A)=cot⁡(A)tan⁡(B)

Step 2: Simplify the equation:

1+tan⁡(A)tan⁡(B)1+tan⁡(A)tan⁡(B)=cot⁡(A)tan⁡(B)

Step 3: The left side of the equation simplifies to 1:

1=cot⁡(A)tan⁡(B)

Step 4: Substitute cot⁡(A)=1tan⁡(A) in the equation:

1=1tan⁡(A)tan⁡(B)

Step 5: Simplify the equation:

1=tan⁡(B)tan⁡(A)

Step 6: Multiply both sides by tan⁡(A):

tan⁡(A)=tan⁡(B)

So, we have proved the given identity.

\tan(A)=\tan(B)

 

[kavindagk]

cot(A) + tan(B) => 1/tan(A) + tan(B) => [1+ tan(A)tan(B)]/tan(A) ------>(1)

cot(B) + tan(A) => 1/tan(B) + tan(A) => [1+ tan(A)tan(B)]/tan(B) -------(2)


(1)/(2) =----> [1+ tan(A)tan(B)]/tan(A) / [1+ tan(A)tan(B)]/tan(B)
cancel both [1+ tan(A)tan(B)]

--> 1/tan(A) / 1/tan(B)
--> cot(A)/cot(B)
---> cot(A)tan(B) <- 1/cotB) => tan(B)

 

[poopoo]

COTA = 1/ Tan A dala balanna enawa bung

Okkoma Tan walata harawanna. mama me dan haduwa nikan mathakada balanna

thanks, awa
chatgpt, bing aluthen trig function invent karanawa

 

[dilann]

හරි සරලයි..

දකුණු පැත්තේ තියන පදේ එලියට අදින්නේ ලවයෙන්..........

LHS =>
cotATanB ( 1/TanB + 1/CotA ) / ( CotB + TanA)
= cotATanB


හරය ලවය කැපිල එලියට ඇදපු එකා ඉතුරු වෙනවා.............

 

[KrishanL89]

Anuradha oya?