Question

If 0 less than a less than B less than pi by ,sin (A+ B) is equal to 24 by 25 , cos( A - B) equal to 4 by 5 find the value of tan 2a

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Question :-

$\sf{if \: \sin \: (A + B) = \frac{24}{25} \: and \: \cos(A - B) = \frac{4}{5} } \\ \sf{then \: find \: \tan2A}$

Answer :-

$\boxed{\sf{ \tan \: 2A = - \frac{3}{4} }} \:$

To find :-

→ Value of tan 2A

Formula used :-

$\star \: \boxed{\sf{ \tan \: (x + y) = \frac{ \tan \: x \: + \tan \: y}{1 - \tan \: x \tan \: y} }}$

Explanation :-

According to the question,

$\implies \sf{ \sin \: (A + B) = \frac{24}{25} = \frac{P}{H} } \\ \: \sf{by \: pythagoras \: theorem} \\ \\ \rightarrow \sf{ {H}^{2} = {P}^{2} + {B}^{2} } \\ \\ \rightarrow \sf{625 = 576 + {B}^{2} } \\ \\ \rightarrow \sf{ {B}^{2} = 49} \\ \\ \rightarrow \boxed{ \sf{ B \: = 7}} \\ \\ \therefore \sf{ \tan(A + B) = \frac{p}{b} = \frac{24}{7} } \:...eq.(1) \\ \\ \sf{now \: from \: } \\ \\ \implies \sf{ \cos \: (A - B) = \frac{4}{5} = \frac{b}{h} } \\ \\ \tt{by \: pythagoras \: theorem} \\ \\ \rightarrow \sf{ {H}^{2} = {P}^{2} + {B}^{2} } \\ \\ \rightarrow \sf{25 = {P}^{2} + 16} \\ \\ \rightarrow \sf{ {P}^{2} = 9} \\ \\ \rightarrow \boxed{\sf{P = 3}} \\ \\ \therefore \sf{ \tan(A - B) = \frac{P}{B} = -\frac{3}{4} } \: ....eq.(2) \\ \sf{ taking \: negative \: \because \: A\: less \: then \:B }\\ \\ \tt hence \\ \\ \rightarrow \sf{ \tan \: 2A = tan(A + A)} \\ \\ \rightarrow \tan \: 2A = \sf{ \tan(A + A + B - B)} \\ \\ \rightarrow \: \tan \: 2A = \tan \big((A + B) + (A - B) \big) \\ \\ \rightarrow \sf{\tan \: 2A = \frac{ \tan(A + B) + \tan(A - B)}{1 - \tan(A + B) \tan(A - B)} } \\ \\ \sf{now \: from \: eq.(1) \: and \: eq.(2)} \\ \\ \rightarrow \sf{ \tan \: 2A = \frac{ \frac{24}{7} - \frac{3}{4} }{1 + \frac{24}{7}. \frac{3}{4} } } \\ \\ \rightarrow \sf{ \tan \: 2A = \frac{ \frac{96 - 21}{28} }{ \frac{28 + 72}{28} } } \\ \\ \rightarrow \boxed{\sf{ \tan \: 2A = \frac{3}{4} }}$

here P is perpendicular , H is hypotenuse and B is base .

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