Question

QUESTION 3 Using the appropriate identity below, find the value of cos cos( 5 – B).ca (Angles are measured in radians.) Formula Sheet Sum & Difference Identities Half Angle Formulas CON 1 + cos(0) 2 cos(0) 2 sin - + cos(a+B) cos(a) cos(8) – sin(a) sin() cos(a-B) cos(a) cos(8) + sin(a) sin() sin(a+b) sin(a) cos(8) + cos(a) sin(8) sin(a -B) sin(a) cow (8) - cos(a) sin() tan(a)tan(B) tan(a+B) 1 - tan(a)tan (8) tan(a)-tan(8) tan(a-) 1+tan(Q) tan() Power Reduction Formulas tan cos(0) 1 + cos(0) Double Angle Identities cox'(0) - sin(0) con(20) 2 cos(0) - 1 1 - 2 sin (0) sin(20) 2 sin(0) cos(0) tan(28) 2 tan(0) 1-tan' cos”(@) 1 + cos(20) 2 1 - cos(20) 2 sin (0) V3 cos(B) - sin(B) ОА. 2 sin(B) - 3cos(B) B. 2

Answer 1

3 con (FIB) using : COA (Q-B) con(a) cos(B) + sin(x)ain (8) Se, conft-B) = cos(9) con{P) +4in (9. )sin (p) 2 2 CORB + ( sin ß COA (T/3) = ² sin (7/3)= fope COAB + 13 ein 2 ie option (D) is correct

As we have verticel a symptotes x=2 and a=-5 y = -3 Ka so, we have in denominator (value must henezer (2-2)(x + 5) or x2 +5x-02-10 = x² + 3x - 10 and horizontal asymptote is Stnce, for the ass horizontal asymptotes to exist numector and denominotor must have same dequel and - 3 will be the vetio of leading coefficient le flae) -3(Q+a)(x+b) (x-2)(x+5) Now it is ginen intercept of -2 (let it be a) le (x+2) will create zuw in hum. + y = f(x) = 2. 501 so, f(x) = -3(x+ 2)(x+b) (2-2) (x+5) given point oin (-4,2) i-e x=-4 putting in we have 2 - 3(-4+ 2-4+6) (1-2)(- 4+) we have [b=2 so, the required equation of retional function is f(x) = -3'(x +2)(x+2) (x-2XX+5)

csc (a)- seelse) Conca)- sin(a) sinacono Taking LHS- le cos () - sin(x). sin(x) con(x) cos(x) sin(x) concx) sinca sin(x) con(x) sin(a) cos(x) since, = ese(x) sinca) and sec(x) con(x) 7 , we have using it esce) – sec (x) = RHS Hence proved