SSC CHSL 9 Jan 2017 Shift 1 Paper

Question : 61 of 100

Marks: +1, -0

If cot(A + B) = x, then value of x is

Solution:

Expression : cot(A + B) = x
=

{sin(A+B)}/ {cos(A+B)}

{sinAcosB+cosAsinB}/ {cosAcosB−sinAsinB}

Dividing both numerator and denominator by , we get :

{cosAcosB-sinAsinB}/ {sinAsinB }

÷ {\ {sinAsinB+cosAcos B}/{sinAsinB }}

({cosAcosB}/{sinAsinB}-1)

÷ ({cos B}/{sinB}+{cosA}/{sinA})

{cotAcotB−1}/{cotB+cotA}

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