In any - ABC- prove that -b B-c C-tan B-tan C-c C-a A-tan C-tan A-a A-b B-tan A-tan B

B sec B+c sec C/tan B+tan C=c sec C+a sec A/tan C+tan A=a sec A+b sec B/tan A+tan B b sec B+c sec C/tan B+tan C=k sin B sec B+k sin C sec C/tan B+tan C =k sin ...

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Byju's Answer

Standard XII

Mathematics

Sine Rule

In any Δ ABC,...

Question

$In any Δ A B C, prove that : \frac{b s e c B + c s e c C}{t a n B + t a n C} = \frac{c s e c C + a s e c A}{t a n C + t a n A} = \frac{a s e c A + b s e c B}{t a n A + t a n B}$

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Solution

$\frac{b s e c B + c s e c C}{t a n B + t a n C} = \frac{c s e c C + a s e c A}{t a n C + t a n A} = \frac{a s e c A + b s e c B}{t a n A + t a n B} \frac{b s e c B + c s e c C}{t a n B + t a n C} = \frac{k s i n B s e c B + k s i n C s e c C}{t a n B + t a n C} = \frac{k s i n B \frac{1}{c o s B} + k s i n C \frac{1}{c o s C}}{t a n B + t a n C} = \frac{k s i n B + k t a n C}{t a n B + t a n C} = \frac{k (t a n B + t a n C)}{t a n B + t a n C} = k Similarly, \frac{c s e c C + a s e c A}{t a n C + t a n A} = k Similary, \frac{a s e c A + b s e c B}{t a n A + t a n B} = k$

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Similar questions

Q. In ∆ABC, prove that:

\frac{b \sec B + c \sec C}{\tan B + \tan C} = \frac{c \sec C + a \sec A}{\tan C + \tan A} = \frac{a \sec A + b \sec B}{\tan A + \tan B}

Q. Assertion :If triangle ABC is equilateral then

t a n A + t a n B + t a n C = 3 \sqrt{3}

Reason: In

Δ A B C

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

If $c o s 2 B = \frac{c o s (A + C)}{c o s (A - C)}$ then

In a triangle tan A + tan B + tan C =

Q. If the sides

a, b, c

of any triangle be in

G . P .

, then prove that

x, y, z

are also in

G . P .

where

x = (b^{2} - c^{2}) \frac{tan B + tan C}{tan B - tan C}

y = (c^{2} - a^{2}) \frac{tan C + tan A}{tan C - tan A}

and

z = (a^{2} - b^{2}) \frac{tan A + tan B}{tan A - tan B}

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In any ΔABC, prove that :b sec B+c sec Ctan B+tan C=c sec C+a sec Atan C+tan A=a sec A+b sec Btan A+tan B

$In any Δ A B C, prove that : \frac{b s e c B + c s e c C}{t a n B + t a n C} = \frac{c s e c C + a s e c A}{t a n C + t a n A} = \frac{a s e c A + b s e c B}{t a n A + t a n B}$