If A- B and C are the angles of a - ABC- prove that tanC-A-2-cotB-2-

If A, B and C are the angles of a ΔABC then the sum of angles is 180° A+B+C= π A+C=π B L. H. S = tan(C+A/2)=tan(π B/2) =tan(π/2 B/2) =tan(90 B/2) =cotB/2= R. ...

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Standard X

Mathematics

Complementary Trigonometric Ratios

If A, B and C...

Question

If A, B and C are the angles of a $Δ A B C,$ prove that $t a n (\frac{C + A}{2}) = c o t \frac{B}{2}$ .

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Solution

If A, B and C are the angles of a ΔABC then the sum of angles is 180°

$A + B + C = π$

$A + C = π - B$

$L . H . S = t a n (\frac{C + A}{2}) = t a n (\frac{π - B}{2})$

$= t a n (\frac{π}{2} - \frac{B}{2})$

$= t a n (90 - \frac{B}{2})$

$= c o t \frac{B}{2} = R . H . S$

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If A, B and C are the angles of a ΔABC, prove that tan(C+A2)=cotB2.

If A, B and C are the angles of a ΔABC then the sum of angles is 180° A+B+C=π A+C=π−B L.H.S=tan(C+A2)=tan(π−B2) =tan(π2−B2) =tan(90−B2) =cotB2=R.H.S

If A, B and C are the angles of a $Δ A B C,$ prove that $t a n (\frac{C + A}{2}) = c o t \frac{B}{2}$ .

If A, B and C are the angles of a ΔABC then the sum of angles is 180°

$A + B + C = π$

$A + C = π - B$

$L . H . S = t a n (\frac{C + A}{2}) = t a n (\frac{π - B}{2})$

$= t a n (\frac{π}{2} - \frac{B}{2})$

$= t a n (90 - \frac{B}{2})$

$= c o t \frac{B}{2} = R . H . S$