If \( \cos B=\frac{p}{\sqrt{5}} \) Whare \( p<0 \) and \( B E\left(180^{\circ}, 360^{\circ}\right) \) Determince using a diagriam an expression in terms of \( P \) for: (a) \( \operatorname{Tan} B \) (b) \( \cos ^{2} B-\sin ^{2} B \) (c) \( \cos \left(180^{\circ}+B\right) \)

We are given that   cos B = p/√5  with p < 0 and that B is in the interval 180° < B < 360°. Notice that in this range cosine is negative. (In fact, cosine is negative in quadrant II and III; here, since B is between 180° and 360°, it must be in quadrant III where both cosine and sine are negative.) Below is one way to “see” the answer by drawing a right–triangle diagram (for the reference angle) and then assigning the proper signs. ────────────────────────────── Step 1. Drawing the Diagram Imagine the terminal side of angle B in quadrant III. Drop a perpendicular from the point on the terminal side to the horizontal axis. The resulting right triangle (the reference triangle) has:  • an adjacent side (along the x–axis) of length |p| (since cos B = p/√5 and p < 0, we use the absolute value for the triangle’s side but then assign the negative sign later for the cosine);  • a hypotenuse of length √5;  • an opposite side, whose length by the Pythagorean theorem is √( (√5)² − (|p|)²) = √(5 − p²). Since B is in quadrant III both the x–coordinate (adjacent) and the y–coordinate (opposite) are negative. Thus,  • cos B = (–|p|)/√5 = p/√5   (since p is already negative),  • sin B = –√(5 – p²)/√5. ────────────────────────────── Now we can find the requested expressions in terms of p. ────────────────────────────── (a) Find tan B By definition,  tan B = sin B / cos B. Substitute the expressions we have:  tan B = (–√(5 – p²)/√5) / (p/√5)      = –√(5 – p²)/p. Thus,  tan B = –√(5 – p²)/p.                    (Answer (a)) ────────────────────────────── ────────────────────────────── (b) Find cos²B – sin²B One method is to recall that  cos²B – sin²B = 2 cos²B – 1     (a standard double–angle formula) We know cos B = p/√5, so  cos²B = (p²)/5. Thus,  cos²B – sin²B = 2(p²/5) – 1 = (2p² – 5)/5.                    (Answer (b)) Alternatively, you could recognize that cos²B – sin²B is cos 2B, but the answer in terms of p is as above. ────────────────────────────── ────────────────────────────── (c) Find cos(180° + B) A standard trigonometric identity tells us that for any angle (or B),  cos(180° + B) = –cos B. Since cos B = p/√5, it follows that  cos(180° + B) = –(p/√5) = –p/√5.                    (Answer (c)) ────────────────────────────── Thus, our final answers are:  (a) tan B = –√(5 – p²)/p  (b) cos²B – sin²B = (2p² – 5)/5  (c) cos(180° + B) = –p/√5 These answers are expressed in terms of p, with the understanding that in the diagram the lengths are taken as positive and then the appropriate negative signs are attached for the quadrant III position.