Question

if sinQ+cosQ=m and secQ+cosecQ=n,then prove that n(m^2-1)2m​

Answer 1

ANSWER

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LHS : n(m2 - 1)

= secA +cosecA [ (sinA +cosA)2 - 1]

= secA + cosecA [ sin2A + cos2A + 2 sinA .cosA -1]

= secA +cosecA [ 1 +2 sinA . cosA -1] [sin2A +cos2A =1]

= secA +cosecA [ 2sinA cosA]

= 2 secA sinA cosA + 2 cosecA sinA cosA

= 2sinA + 2 cosA [secA*cosA=1 and cosecA*sinA=1]

= 2(sinA +cosA)

= 2 m [sinA +cosA =m]

= RHS

Hope helps friend.......

Answer 2

Given, sin Q + cos Q = m

Squaring both sides of the above equation, we will get :

1 + 2 sin Q cos Q = m²

Now, LEFT HAND SIDE :

n (m² - 1)

= (sec Q + cosec Q) (1+2 sin Q cos Q -1)

= ((1/cos Q) + (1/sin Q)) (2 sin Q cos Q)

= 2 (sin Q + cos Q)

= 2 m [ PUTTING THE VALUE OF sin Q + cos Q ]

= RIGHT HAND SIDE

THEREFORE, L.H.S. = R.H.S. (PROVED)