Question

If 4 cot A=3,then find (sinA+cosA)sinA-cosA)

Answer 1

⚝ Given :-

• 4 cot A = 3

⚝ Find :-

• (sinA+cosA) (sinA-cosA)

⚝ Solution :-

$\leadsto{ \sf{4 \: cot A = 3}}$

$\leadsto{ \sf{ cot A = \frac{3}{4} }}$

• As we know that,

$\leadsto{ \sf{cot A = \frac{B}{P} }}$

• We can say that,

$\leadsto{ \sf{\frac{B}{P} = \frac{3}{4} }}$

• Thus,

• Base (B) = 3
• Perpendicular (P) = 4

• CALCULATING HYPOTENUSE :

By Using Pythagoras Theorem :-

⇝ H² = B² + P²

⇝ H² = 3² + 4²

⇝ H² = 9 + 16

⇝ H² = 25

⇝ H = √25

⇝ H = 5

• Therefore, Hypotenuse is 5. Now,

$\leadsto{ \boxed{ \sf{sin A = \frac{P}{H} }}}$

• Substitute the values :

$\leadsto{ \boxed{ \sf{sin A = \frac{4}{5} }}} \: \bigstar$

• Similarly,

$\leadsto{ \boxed{ \sf{cos A = \frac{B}{H} }}}$

• Substitute the values :

$\leadsto{ \boxed{ \sf{cos A = \frac{3}{5} }}} \: \bigstar$

• Now,

$\leadsto \displaystyle{ \sf{ \frac{(sinA+cosA) }{(sinA-cosA)}}}$

$\leadsto \displaystyle{ \sf{ \frac{ \frac{4}{5} + \frac{3}{5} }{ \frac{4}{5} - \frac{3}{5} } }}$

$\leadsto \displaystyle{ \sf{ \frac{ \frac{7}{5} }{ \frac{1}{5} } }}$

$\leadsto \displaystyle{ \sf{ \frac{7}{5} \times \frac{5}{1} }}$

$\leadsto \displaystyle{ \sf{ \frac{7}{ \cancel5} \times \frac{ \cancel5}{1} }}$

$\leadsto \displaystyle{ \sf{ \frac{7}{1} = 7}}$

• Henceforth, the value of $\displaystyle{ \sf{ \frac{(sinA+cosA) }{(sinA-cosA)}}}$ is 7.

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