If A - B - C be the angles of a triangle- then - A - B -tan A -tan B -A- 1B- 2C- -1D- -2

The correct option is A 1∑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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Conditional Identities

If A , B , C ...

Question

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

Similar questions

If in a right-angled triangle ABC angles A and B are acute, then evaluate

1 + $\frac{t a n A}{t a n B}$ =

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

If in a right-angled triangle ABC angles A and B are acute, then evaluate

1 + $\frac{t a n A}{t a n B}$ =

Q. If

A,

B

and

C

are the angles of a non-right angled triangle

A B C,

then the value of

∣ ∣ ∣ ∣ \begin{matrix} tan A & 1 & 1 1 & tan B & 1 1 & 1 & tan C \end{matrix} ∣ ∣ ∣ ∣

is equal to

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}

Join BYJU'S Learning Program

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