sec B=(-17/8), pi<B<(3pi/2) find sin B, cos B, and tan B....help please!

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sec B = -17/8 and pi < B < 3pi/2

In the given range of B, both sin and cos have negative values and therefore, tan has a positive value.

now, sec is inverse of cos.

Thus, sec B = 1/cosB = -17/8

means, cos B = -8/17

sin B = `sqrt (1-cos^2B) = sqrt(1-(-8/17)^2) = - 15/17`  

and tan B = sinB/cosB = `(-15/17)/(-8/17) = 15/8`

Hope this helps.

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