If A- B- C be the angles of a triangle- then -cot A-cot B-tan A-tan B-

∑cot A+cot B/tan A+tan B =∑cos A/sin A+cos B/sin B/sin A/cos A+sin B/cos B =∑sin B cos A+sin A cos B/sin A. sin B*cos A. cos B/(sin A cos B+cos A. sin B) =∑ cot ...

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Standard XI

Mathematics

Conditional Identities

If A, B, C be...

Question

If A, B, C be the angles of a triangle, then $\sum \frac{c o t A + c o t B}{t a n A + t a n B} =$

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Solution

The correct option is A 1
$\sum \frac{c o t A + c o t B}{t a n A + t a n B} = \sum \frac{\frac{c o s A}{s i n A} + \frac{c o s B}{s i n B}}{\frac{s i n A}{c o s A} + \frac{s i n B}{c o s B}}$
= $\sum \frac{s i n B c o s A + s i n A c o s B}{s i n A . s i n B} * \frac{c o s A . c o s B}{(s i n A c o s B + c o s A . s i n B)}$
= $\sum c o t A c o t B$
As we know if $A + B + C = π$ , then
cot A cot B + cot B cot C + cot C cot A = 1.

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Similar questions

Q. If

A, B, C

are angles of triangle then
(i)

tan A + tan B + tan C = tan A tan B tan C

(ii)

cot A cot B + cot B cot C + cot C cot A = 1

Q. If

A, B, C

are the angles of a triangle and if none of them is equal to

\frac{π}{2}

, then find

tan A + tan B + tan C - tan A tan B tan C

+

cot A cot B + cot B cot C + cot C cot A

Q. If A, B, C are the angles of a triangle, show that

\frac{cot A + cot B}{tan A + tan B} + \frac{cot B + cot C}{tan B + tan C} + \frac{cot C + cot A}{tan C + tan A} = 1

Q. If A, B, C be the angles of a triangle, then

\sum \frac{c o t A + c o t B}{t a n A + t a n B} =

I f A, B, C

be the angle of a triangle, the

\sum \frac{c o t A + c o t B}{t a n A + t a n B}

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If A, B, C be the angles of a triangle, then ∑cot A+cot Btan A+tan B=

The correct option is A 1∑cot A+cot Btan A+tan B=∑cos Asin A+cos Bsin Bsin Acos A+sin Bcos B =∑sin B cos A+sin A cos Bsin A. sin B∗cos A. cos B(sin A cos B+cos A. sin B) =∑cot A cot B As we know if A+B+C=π, then cot A cot B + cot B cot C + cot C cot A = 1.

If A, B, C be the angles of a triangle, then $\sum \frac{c o t A + c o t B}{t a n A + t a n B} =$