If cosecA - A-cosec B- B- provethat - tan A tan B-A-B-2

We have, cosecA+secA=cosecB+secB secA secB=cosecB cosecB ⇒1/cos A 1/cos B=1/sin B 1/sin A ⇒cosB cosA/cosA cosB=sinA sinB/sinA sinB ⇒ tanA tanB =2sin ( A B/2 ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

If cosecA + A...

Question

If $c o s e c A + s e c A = c o s e c B + s e c B, p r o v e t h a t : t a n A t a n B = c o t \frac{A + B}{2}$

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Solution

We have,
$c o s e c A + s e c A = c o s e c B + s e c B s e c A - s e c B = c o s e c B - c o s e c B \Rightarrow \frac{1}{c o s A} - \frac{1}{c o s B} = \frac{1}{s i n B} - \frac{1}{s i n A} \Rightarrow \frac{c o s B - c o s A}{c o s A c o s B} = \frac{s i n A - s i n B}{s i n A s i n B} \Rightarrow t a n A t a n B = \frac{2 s i n (\frac{A - B}{2}) C O S (\frac{A + B}{2})}{- 2 s i n (\frac{B - A}{2}) s i n (\frac{B + A}{2})} \Rightarrow t a n A t a n B = \frac{- s i n (\frac{A - B}{2}) c o s (\frac{A + B}{2})}{- s i n (\frac{A - B}{2}) s i n (\frac{A + B}{2})} [∵ s i n (- θ) = - s i n θ] \Rightarrow t a n A t a n B = c o t (\frac{A + B}{2})$

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Similar questions

Q. If cosec A + sec A = cosec B + sec B, prove that tan A tan B =

\cot \frac{A + B}{2}

Q. If cosec (A + B) = 1 and cosec(A – B) = 2, 0° < (A + B) ≤ 90° and A > B then find the values of :
(i) sin A cos B + cos A sin B
(ii)

\frac{\tan A - \tan B}{1 + \tan A \tan B}

Q. In

△ A B C

c o s e c \frac{A}{2} + c o s e c \frac{B}{2} + c o s e c \frac{C}{2} \geq m

Find m

Q. If

cosec A + cosec B + cosec C = 0

, then show that

{(sin A + sin B + sin C)}^{2} = \sum {sin}^{2} A

(TanA +cosec B)^2-(cot B-sec A)^2=2tanAcotB(cosecA+secB)

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If cosecA+secA=cosecB+secB,provethat:tanA tanB=cotA+B2

We have, cosecA+secA=cosecB+secBsecA−secB=cosecB−cosecB⇒1cosA−1cosB=1sinB−1sinA⇒cosB−cosAcosA cosB=sinA−sinBsinA sinB⇒tanA tanB=2sin(A−B2)COS(A+B2)−2sin(B−A2)sin(B+A2)⇒tanA tanB=−sin(A−B2)cos(A+B2)−sin(A−B2)sin(A+B2)[∵sin(−θ)=−sinθ]⇒tanA tanB=cot(A+B2)

If $c o s e c A + s e c A = c o s e c B + s e c B, p r o v e t h a t : t a n A t a n B = c o t \frac{A + B}{2}$