If sin B = 15⁄17, what is the value of cos B(sec B − tan B)? Given that 0 < B < π⁄2

Correct — B. Multiply the bracket out before any number goes in.

cos B × sec B = 1, since the two are reciprocals
cos B × tan B = cos B × (sin B / cos B) = sin B

So cos B(sec B − tan B) = 1 − sin B

= 1 − 15/17
= 17/17 − 15/17 = 2/17 → option (b)

cos B never has to be worked out at all — the bracket disposes of it.

Why the others are wrong

• (a)
  8/15 is cot B. From the 8-15-17 right triangle cos B = 8/17 and sin B = 15/17, so cos B / sin B = 8/15 — a correct value, answering a question the stem does not ask.
• (c)
  1/16 relies on a 16 that does not exist here. The sides are 8, 15 and 17, and 1 − 15/17 carries the denominator 17, so no step in this working produces a sixteenth.
• (d)
  2/15 has the numerator right and the denominator wrong. 1 − 15/17 = (17 − 15)/17, so the 2 sits over the hypotenuse 17, not over the opposite side 15.

Simplify before you substitute. cos B(sec B − tan B) expands to cos B·sec B − cos B·tan B, and each product collapses: cos B·sec B = 1 and cos B·tan B = sin B.

So the expression is 1 − sin B for every B where the ratios are defined. Only then does sin B = 15/17 matter, and it gives 2/17.

The stated range 0 < B < π/2 keeps cos B positive, but the simplified form never uses cos B, so the value would be the same without it.

Key facts

• cos B and sec B are reciprocals, so their product is 1 wherever sec B is defined.
• cos B × tan B = sin B, because tan B is sin B divided by cos B.
• 8, 15, 17 is a Pythagorean triple, so an acute B with sin B = 15/17 has cos B = 8/17.

Common traps

• Working out cos B = 8/17 first and then hunting for somewhere to use it
• Reading cos B(sec B − tan B) as cos B minus the bracket
• Writing 1 − 15/17 as 2/15

No directly related past PYQ was found.