Question

If 5SinA + 3CosA = 4, find the value of 3sinA – 5Cos A.
Class 10 Maths Trigonometry

Answer 1

Answer:

Step-by-step explanation:

5 Sin A+ 3Cos A= 4 ----> eq 1

3 Sin A- 5 Cos A = x -----> eq 2

squaring and adding both the equations:

(5 sin A+3 cos A)²+ (3sin A-5Cos A)² = 4²+x²

25 sin²A+ 2×3×5 SinA cos A+ 9cos²A + 9²sinA - 2×3×5 sinA cos A + 25 Cos²A = 4² + x²

rearranging :

25 sin ²A + 25Cos²A+9 sin²A +9cos²A+30 sinAcosA-30SinAcosA= 16+x²

(since sin²A+cos²A = 1)

25+9 = 16+x²

x² = 34

x =4.242

Answer 2

Answer:

$\begin{gathered}\sf If \: 3sinA+5cosA=5 \:then \\\\\sf 5sinA-3cosA=±3 \end{gathered}$

Step-by-step explanation:

Given \: 3sinA+5cosA=5Given3sinA+5cosA=5

• On Squaring both sides of the equation, we get

$\sf \implies \left(3sinA+5cosA\right)^{2}=5^{2}$

$\sf\implies (3sinA)^{2}+(5cos)^{2}-2\times 3sinA\times 5cosA=25⟹(3sin)$

* By algebraic identity:*

• (a+b)² = a²+b²+2ab */

$\sf\implies 9sin^{2}A+25cos^{2}A+30sinAcosA=25$

$\sf\implies 9(1-cos^{2}A)+25(1-sin^{2}A)$

$\sf\implies 9-9cos^{2}A+25-25sin^{2}A+30sinAcosA=25$

$\sf\implies 9=9cos^{2}A+25sin^{2}$

Therefore,

• 5sinA-3cosA = ±35sinA−3cosA=±3

•••