Question

Pls find the question below in the attachment.

Answer 1

QuestioN :

If Sin A = 3/5, 0 < A < π/2 and Cos B = - 12 /13 , π < B < 3π/2 l, then find the following l.

( i ) Sin( A - B )

( ii ) Cos( A + B )

( iii ) tan( A - B )

AnsWer :

( i ) Sin( A - B ) = - 16/65.

( ii ) Cos( A + B ) = - 33/65.

( iii ) tan( A - B ) = 16/63.

SolutioN :

Taking Sin A.

We know,

• Sin² A + Cos² A = 1.

So, Let's Find the value of Cos A.

$\tt \dagger \: \: \: \: \: \fbox{ {sin}^{2} A + {Cos}^{2} A = 1.}$

$\tt : \implies {Cos}^{2} A = 1 - {sin}^{2} A$

$\tt : \implies {Cos}A = \sqrt{ 1 - {sin}^{2} A}$

$\tt : \implies Cos A = \sqrt{ 1 - { \bigg(\dfrac{3}{5} \bigg) }^{2} }$

$\tt : \implies Cos A = \sqrt{ 1 - {\dfrac{9}{25} } }$

$\tt : \implies Cos A = \sqrt{ {\dfrac{ 25 - 9}{25} } }$

$\tt : \implies Cos A = \sqrt{ {\dfrac{16}{25} } }$

$\tt : \implies Cos A = {\dfrac{4}{5} }$

We have,

• Cos B = - 12 / 13.

Let Find the value of Sin B.

$\tt : \implies {sin}B = \sqrt{ 1 - {cos}^{2} B}$

$\tt : \implies {sin}B = - \sqrt{ 1 - {\bigg(- \dfrac{12}{13} \bigg) }^{2} }$

$\tt : \implies {sin}B = - \sqrt{ 1 - \bigg(- \dfrac{144}{169} \bigg) }$

$\tt : \implies {sin}B = - \sqrt{\dfrac{169 - 144}{169} }$

$\tt : \implies {sin}B = - \sqrt{\dfrac{169 - 144}{169} }$

$\tt : \implies {sin}B = - \dfrac{5}{13}$

$\rule{120}3$

★ Now,

Let's Find the value of

( i ) Sin( A - B )

We have Formula.

• Sin( A - B ) = Sin A. Cos B - Cos A . Sin B.

$\tt : \implies Sin( A - B ) = \dfrac{3}{5} \times \dfrac{12}{13} - \dfrac{4}{5} \times - \dfrac{5}{13}$

$\tt : \implies Sin( A - B ) = - \dfrac{36}{65} + \dfrac{20}{65}$

$\tt : \implies Sin( A - B ) = - \dfrac{16}{65}$

$\rule{120}3$

( ii ) Cos( A - B )

We have Formula.

• Cos( A - B ) = Cos A. Cos B - Sin A . Sin B.

$\tt : \implies Cos( A + B ) = \dfrac{4}{5} \times - \dfrac{12}{13} - \dfrac{3}{5} \times - \dfrac{5}{13}$

$\tt : \implies Cos( A + B ) = - \dfrac{48}{65} + \dfrac{15}{65}$

$\tt : \implies Cos( A + B ) = - \dfrac{33}{65}$

$\rule{120}3$

( iii ) Tan( A - B )

We have Formula.

• tan A - tan B / 1 + tan A . tan B.

$\tt : \implies tan A = \dfrac{ \frac{3}{5} }{ \frac{4}{5} } = \dfrac{3}{4}$

$\tt : \implies tan B = \dfrac{ - \frac{5}{13} }{ - \frac{12}{13} } = \dfrac{5}{12}$

Now,

$\tt : \implies tan (A + B) = \frac{ \frac{3}{4} - \frac{5}{12} }{1 + \frac{3}{4} \times \frac{5}{12} }$

$\tt : \implies tan (A + B) = \dfrac{16}{63}$

Note : More information attached above.