Class11 Maths NCERT Solutions. Class 12 Maths NCERT Solutions. Class 8 Science NCERT Solutions. Class 9 Science NCERT Solutions. Class 10 Science NCERT Solutions. Englishtan. NCERT Books. Class 10 Maths - Basic vs Standard. CBSE Sample Papers. If \[\sin A = \frac{4}{5}\] and \[\cos B = \frac{5}{13}\], where 0 < A, \[B < \frac{\pi}{2}\], find the value of the followingsin A − BGiven \[ \sin A = \frac{4}{5}\text{ and }\cos B = \frac{5}{13}\]We know that\[ \cos A = \sqrt{1 - \sin^2 A}\text{ and }\sin B = \sqrt{1 - \cos^2 B} ,\text{ where }0 < A , B < \frac{\pi}{2}\]\[ \Rightarrow \cos A = \sqrt{1 - \left \frac{4}{5} \right^2} \text{ and }\sin B = \sqrt{1 - \left \frac{5}{13} \right^2}\]\[ \Rightarrow \cos A = \sqrt{1 - \frac{16}{25}}\text{ and }\sin B = \sqrt{1 - \frac{25}{169}}\]\[ \Rightarrow \cos A = \sqrt{\frac{9}{25}}\text{ and }\sin B = \sqrt{\frac{144}{169}}\]\[ \Rightarrow \cos A = \frac{3}{5}\text{ and }\sin B = \frac{12}{13}\]Now,\[\sin\left A - B \right = \sin A \cos B - \cos A \sin B \]\[ = \frac{4}{5} \times \frac{5}{13} - \frac{3}{5} \times \frac{12}{13}\]\[ = \frac{20}{65} - \frac{36}{65}\]\[ = \frac{- 16}{65}\]
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Q Find the exact value of cos(A+B) if sin A = with A in QI and sin B 5 with B in QID = - A: Given: sinA=35sinB=-513 To find the exact value of cos(A+B) COS(A+B)=COSACOSB-SINASINB COS2θ+SIN2θ=1 Q: find sin ). co ), aud tan for each gueton Sin X = 4 and x is ふ quadrant
The correct option is D-1665Explanation for the correct 1 Find the value of cosA,sinBGiven that, sinA=45and cosB= know that, sin2θ+cos2θ=1cosA=1-sin2A=1-452=35Now the value of sinBis negative because B lies in 3rd quadrant. sinB=1-12132=1-144169=25169=-513Step 2 Find the value of cosA+BWe know that, cosA+B= option D is correct. 43 5) cos A 30 16 34 A B C 15 17 6) sin A 24 32 40 A C B 4 5 7) sin Z 32 24 40 Z Y X 3 5 8) sin C 48 14 50 C B A 7 25 cos C 36 27 45 C B A 0.6000 13) tan C 40 30 50 C B A 1.3333 14) tan A 21 20 29 A B C 1.0500 15) tan C 35 12 37 B C A 0.3429 16) tan X 40 30 X 50 Y Z 0.7500 17) sin Z 35 12 37 ZY X 0.3243 18) sin Z >>Class 11>>Maths>>Trigonometric Functions>>Trigonometric Functions of Sum and Difference of Two angles>>If cos A = 4/5 , cos B = 12/13 , 3pi/Open in AppUpdated on 2022-09-05SolutionVerified by TopprA and B both lie in the IV quadrant.=> are negativei iiSolve any question of Trigonometric Functions with-Was this answer helpful? 00More From ChapterLearn with Videos Practice more questions Ifsin A = 4/5 and cos B = 5/13, where 0 . CBSE | Class 11 | Exercise 7.1 | Maths | RD Sharma | Values of Trigonometric Functions at Sum or Difference of Angles | Class 11 | Exercise 7.1 | Maths | RD Sharma | Values of Trigonometric Functions at Sum or Difference of Angles Here, colorgreenI^st Quadrant=> 0 all+ve sina=5/13=>cosa=sqrt1-sin^2a=sqrt1-25/169=12/13 cosb=4/5=>sinb=sqrt1-cos^2b=sqrt1-16/25=3/5 colorredisina+b=sinacosb+cosasinb colorwhiteisina+b=5/13xx4/5+12/13xx3/5=20/65+36/65=56/65 colorblueiicosa-b=cosacosb+sinasinb colorwhiteiicosa-b=12/13xx4/5+5/13xx3/5=48/65+15/65=63/65 colorvioletiiicosb/2=sqrt1+cosb/2=sqrt1+4/5/2=sqrt9/10=3/sqrt10 colororangeivsin2a=2sinacosa=2xx5/13xx12/13=120/169 I've used the angle sum identity to end up with $\cos A \cos B -\sin A \sin B = \frac{5}{13} = \frac{3}{5}\cos B -\frac{4}{5} \sin B$, but don't know how to proceed from here.
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
Ifthe sine or cosine function has a coefficient of one, isolate the term on one side of the equals sign. If the number it is set equal to has an absolute value less than or equal to one, the equation has solutions, otherwise it does not. If the sine or cosine does not have a coefficient equal to one, still isolate the term but then divide both If sin A = 4 5 and cos B =-12 13, where A and B lie in the first and third quadrant respectively, then cos (A + B) = A. 56 65. No worries! We've got your back. Try BYJU'S free classes today! B-56 65. No worries! We've got your back. Try BYJU'S free classes today! C. 16 65. No worries! We've got your back. Try BYJU'S free classes today! FreeOnline Scientific Notation Calculator. Solve advanced problems in Physics, Mathematics and Engineering. Math Expression Renderer, Plots, Unit Converter, Equation Solver, Complex Numbers, Calculation History. DoCeSdR.
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