Q.336. Trigonometric proof.

Question 336.

If sin2x = 22 - 8√7,  prove that sinx + cosx + tanx + cosecx + secx + cotx = 7, given that angle x is acute.

Answer 336.

I had not answered this question, but while voting, I found this question to be good and have a better simpler answer posted here.

sin2x = 22 - 8√7

=> 2sinx cosx = 22 - 8√7

=> sinx cosx = 11 - 4√7   ...   ( 1 )

1 + sin2x = 23 - 8√7

=> (sinx + cosx)^2 = 23 - 2√(112) = (4 - √7)^2

=> sinx + cosx = 4 - √7   ...   ( 2 )

sinx + cosx + tanx + cosecx + secx + cotx

= (sinx + cosx) + (cosecx + secx) + (tanx + cotx)

= (sinx + cosx) + (1/sinx + 1/cosx) + (sinx/cosx + cosx/sinx)

= (sinx + cosx) + (sinx + cosx) / sinx cosx + (sin^2 x + cos^2 x) / sinx cosx

= (4 - √7) + (4 - √7) / (11 - 4√7) + 1 / (11 - 4√7)

= [(4 - √7) * (11 - 4√7) + (4 - √7) + 1] / (11 - 4√7)

= (44 + 28 - 27√7 + 5 - √7) / (11 - 4√7)

= 7 * (11 - 4√7) / (11 - 4√7)

= 7.

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