Inthe triangle ABC right-angled at B, if tan A = 1/√3 find the value of: (i) sin A cos C + cos A sin C (ii) cos A cos C - sin A sin C. We use the trigonometric ratios to solve the problem. In the triangle ABC right-angled at B, if tan A = 1/√3 then the value of sin A cos C + cos A sin C = 1, and cos A cos C - sin A sin C = 0. Thefirst shows how we can express sin θ in terms of cos θ; the second shows how we can express cos θ in terms of sin θ. Note: sin 2 θ-- "sine squared theta" -- means (sin θ) 2. Problem 3. A 3-4-5 triangle is right-angled. a) Why? To see the answer, pass your mouse over the colored area. To cover the answer again, click "Refresh" ("Reload"). Findthe exact value of sin( a−B), given that sin a=−4/5 and cos B=12/13, with a in quadrant III and B in quadrant IV. 2. Find all real numbers in the interval [0,2pi) that satisfy the equation. 3sec^2 x tan x =4tan x 3. Simplify the following trigonometric expressions, using identities as needed: Vay Tiền Trả Góp Theo Tháng Chỉ Cần Cmnd Hỗ Trợ Nợ Xấu. We have, sinA=45 and B=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1−452 and sinB=√1−5132sinB=√1−5132 ⇒cosA=√1−1625 and sinB=√1−25169 ⇒cosA=√25−1625 and sinB=√169−25169 ⇒cosA=√925 and sinB=√144169⇒cosA=35 and sinB=1213 Now, sinA+B=sinA cosB+cosA sinB =45×513+35×1213=2065+3665=20+3665=5665 ii We have, sinA=45 and B=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1452 and sin B=√1−5132 ⇒cosA=√1−1615 sin B=√1−25169 ⇒cosA=√25−1625 and sin B=√169−25169 ⇒cosA=√925 and sin B=√144169 cosA=35 and sinB=1213 Now, cosA+B=cosA cosB−sinA sinB =35×513−45×1213 =1565−4865 =15−4865=−3365 iii We have, sinA=45 and cosB=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1−452 and sinB=√1−5132 ⇒cosA=√1−1625 and sinB=√1−25169 ⇒cosA=√25−1625 and sinB=√169−25169 ⇒cosA=√925 and sinB=√144169 ⇒cosA=35 and sinB=1213 Now, sinA−B=sinA cosB−cosA sinB =45×513−35×1213 =2065+3665=20−3665=−1665 iv We have, sinA=45 and cosB=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1−452 sinB=√1−5132 ⇒cosA=√1−1625 and sinB=√1−25169 ⇒cosA=√25−1625 and sinB=169−25169 ⇒cosA=√925 and sinB=1213 Now, cosA−B=cosA cosB+sinA sinB =35×513+45×1213 =1565+4865 =15+4865 =6365 given, cosA+B=4/5, thus tanA+B=3/4 from triangle sinA-B=5/13,thus tanA-B=5/12. then tan2A=tanA+B+A-B =tanA+B+tanA-B/1-tanA+BtanA-B =3/4+5/12/1-3/45/12 = 56/33. Open in Appwe have the value of and but we don't have the value of and so, first we find the value of and let side opposite to angle hypotenuse where is any positive integer So, by Pythagoras theorem we can find the third side of a triangle taking positive square root as side cannot be negative So, Base we know that side adjacent to angle hypotenuse so, now we have to find the we know that let side adjacent to angle hypotenuse where is any positive integer so, by Pythagoras theorem, we can find the third side of a triangle taking positive square root since, side cannot be negative so, perpendicular we know that Now putting the values, we get Was this answer helpful? 00

sin a 4 5 cos b 5 13