CameraIcon

SearchIcon

MyQuestionIcon

1

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

Solving a Quadratic Equation by Factorization Method

Tan A+cosec B...

Question

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

Open in App

Solution

LHS
= (tanA + cosecB)^2 - (cotB-secA)^2
= (tan^2A + 2tanAcosecB + cosec^2B) - (cot^2B - 2cotBsecA + sec^2A)
= tan^2A + cosec^2B + 2tanAcosecB - cot^2B - sec^2A + 2cotBsecA

Substituting (tan^2A = sec^2A - 1) and (cosec^2B = 1 + cot^2B) in the above step:
(sec^2A - 1) + (1 + cot^2B) + 2tanAcosecB - cot^2B - sec^2A + 2cotBsecA
= 2tanAcosecB + 2cotBsecA
= 2(tanAcosecB + cotBsecA)

RHS
= 2tanAcotB(cosecA + secB)
= 2tanAcotB((1 / sinA) + (1 / cosB))
= 2(tanA / sinA)cotB + 2tanA(cot B / cosB)
= 2(sinA / (cosAsinB))cotB + 2tanA(cosB / (sinBcosB))
= 2(1 / cosA)cotB + 2 tanA(1 / sinB)
= 2secAcotB + 2tanAcosecB
= 2(secAcotB + tanAcosecB)

Suggest Corrections

thumbs-up

57

Similar questions

Q. If A and B are complementary angles, then

(a) sin A = sin B
(b) cos A = cos B
(c) tan A = tan B
(d) sec A = cosec B

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

$\frac{a - b}{a + b} = \frac{t a n (\frac{A - B}{2})}{t a n (\frac{A + B}{2})}$

Q. Prove that :

\frac{c}{a - b} = \frac{tan \frac{A}{2} + tan \frac{B}{2}}{tan \frac{A}{2} - tan \frac{B}{2}}

$\frac{c}{a + b} = \frac{1 - t a n (\frac{A}{2}) t a n (\frac{B}{2})}{1 + t a n (\frac{A}{2}) t a n (\frac{B}{2})}$

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Solving QE by Factorisation

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Solving a Quadratic Equation by Factorization Method

Standard X Mathematics

Join BYJU'S Learning Program

CrossIcon