Question

If 4sinФ= 3 cosФ, show that 12sinФ - 7cosФ/ 8sinФ + 3cosФ = 3/4 .
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Answer 1

S O L U T I O N : (ques.error)

$\underline{\bf{Given\::}}$

4 sin Ф = 3 cos Ф, show that 12 sinФ - 7 cosФ/8 sinФ + 3 cosФ = 2/9 .

$\underline{\bf{Explanation\::}}$

$\mapsto\tt{ 4 sin \:\theta = 3 cos \:\theta }$

$\mapsto\tt{ \dfrac{sin\:\theta }{cos\:\theta } = \dfrac{3}{4} }$

$\mapsto\tt{ tan \:\theta = \dfrac{3}{4}\:\:\:\bigg[\therefore tan \:\theta = sin\:\theta / cos\:\theta \bigg]}$

Taking L.H.S :

$\mapsto\tt{\dfrac{12 sin\:\theta - 7 cos\:\theta }{8 sin\:\theta + 3 cos\:\theta} }$

$\mapsto\tt{\dfrac{12\frac{sin\:\theta }{cos\:\theta } - 7 \frac{cos\:\theta}{cos\:\theta} }{8 \frac{sin\:\theta}{cos\:\theta} + 3\frac{cos\:\theta}{cos\:\theta} } }$

$\mapsto\tt{\dfrac{12tan\:\theta - 7 }{8tan\:\theta + 3} }$

$\mapsto\tt{\dfrac{12\bigg(\dfrac{3}{4}\bigg) - 7 }{8\bigg(\dfrac{3}{4} \bigg)+ 3} }$

$\mapsto\tt{\dfrac{\dfrac{36}{4} - 7 }{\dfrac{24}{4} + 3} }$

$\mapsto\tt{\dfrac{\cancel{\dfrac{36}{4}} - 7 }{\cancel{\dfrac{24}{4}} + 3} }$

$\mapsto\tt{\dfrac{9 - 7 }{6 + 3} }$

$\mapsto\bf{\dfrac{2}{9} }$

Thus,

L.H.S = R.H.S    [verified]

Answer 2

Answer:

S O L U T I O N : (ques.error)

\underline{\bf{Given\::}}

Given:

4 sin Ф = 3 cos Ф, show that 12 sinФ - 7 cosФ/8 sinФ + 3 cosФ = 2/9 .

\underline{\bf{Explanation\::}}

Explanation:

\mapsto\tt{ 4 sin \:\theta = 3 cos \:\theta }↦4sinθ=3cosθ

\mapsto\tt{ \dfrac{sin\:\theta }{cos\:\theta } = \dfrac{3}{4} }↦

cosθ

sinθ

=

4

3

\mapsto\tt{ tan \:\theta = \dfrac{3}{4}\:\:\:\bigg[\therefore tan \:\theta = sin\:\theta / cos\:\theta \bigg]}↦tanθ=

4

3

[∴tanθ=sinθ/cosθ]

Taking L.H.S :

\mapsto\tt{\dfrac{12 sin\:\theta - 7 cos\:\theta }{8 sin\:\theta + 3 cos\:\theta} }↦

8sinθ+3cosθ

12sinθ−7cosθ

\mapsto\tt{\dfrac{12\frac{sin\:\theta }{cos\:\theta } - 7 \frac{cos\:\theta}{cos\:\theta} }{8 \frac{sin\:\theta}{cos\:\theta} + 3\frac{cos\:\theta}{cos\:\theta} } }↦

8

cosθ

sinθ

+3

cosθ

cosθ

12

cosθ

sinθ

−7

cosθ

cosθ

\mapsto\tt{\dfrac{12tan\:\theta - 7 }{8tan\:\theta + 3} }↦

8tanθ+3

12tanθ−7

\mapsto\tt{\dfrac{12\bigg(\dfrac{3}{4}\bigg) - 7 }{8\bigg(\dfrac{3}{4} \bigg)+ 3} }↦

8(

4

3

)+3

12(

4

3

)−7

\mapsto\tt{\dfrac{\dfrac{36}{4} - 7 }{\dfrac{24}{4} + 3} }↦

4

24

+3

4

36

−7

\mapsto\tt{\dfrac{\cancel{\dfrac{36}{4}} - 7 }{\cancel{\dfrac{24}{4}} + 3} }↦

4

24

+3

4

36

−7

\mapsto\tt{\dfrac{9 - 7 }{6 + 3} }↦

6+3

9−7

\mapsto\bf{\dfrac{2}{9} }↦

9

2

Thus,

L.H.S = R.H.S [verified]