Question

plz zzz help me out.... ​

Answer 1

Answer:-

$\displaystyle\boxed{\red{\sf\:\dfrac{\left(\:2\:\sin\:x\:-\:3\:\cos\:x\:\right)}{\left(\:4\:\sin\:x\:-\:9\:\cos\:x\:\right)}\:=\:3}}$

Step-by-step-explanation:-

We have given that $\sf\:\sec\:x\:=\:\dfrac{13}{5}$

We have to prove that,

$\displaystyle\sf\:\dfrac{\left(\:2\:\sin\:x\:-\:3\:\cos\:x\:\right)}{\left(\:4\:\sin\:x\:-\:9\:\cos\:x\:\right)}\:=\:3$

Now, we know that,

$\displaystyle\pink{\sf\:\sec\:x\:=\:\dfrac{1}{\cos\:x}}\sf\:\:\:-\:-\:-\:[\:Trigonometric\:identity\:]\\\\\\\implies\sf\:\dfrac{13}{5}\:=\:\dfrac{1}{\cos\:x}\\\\\\\implies\sf\:\cos\:x\:=\:\dfrac{1}{\dfrac{13}{5}}\\\\\\\implies\sf\:\cos\:x\:=\:\dfrac{1\:\times\:5}{13}\\\\\\\implies\boxed{\red{\sf\:\cos\:x\:=\:\dfrac{5}{13}}}$

Now, we know that,

$\displaystyle\pink{\sf\:\sin^2\:x\:+\:\cos^2\:x\:=\:1}\sf\:\:\:-\:-\:-\:[\:Trigonometric\:identity\:]\\\\\\\implies\sf\:\sin^2\:x\:+\:\left(\:\dfrac{5}{13}\:\right)^2\:=\:1\\\\\\\implies\sf\:\sin^2\:x\:=\:1\:-\:\left(\:\dfrac{5}{13}\:\right)^2\\\\\\\implies\sf\:\sin\:x\:=\:\sqrt{1\:-\:\left(\:\dfrac{5}{13}\:\right)^2}\:\:\:-\:-\:[\:Taking\:square\:roots\:]\\\\\\\implies\sf\:\sin\:x\:=\:\sqrt{\:\left(\:1\:\right)^2\:-\:\left(\:\dfrac{5}{13}\:\right)^2}\\\\\\\implies\sf\:\sin\:x\:=\:\sqrt{\left(\:1\:+\:\dfrac{5}{13}\:\right)\:\left(\:1\:-\:\dfrac{5}{13}\:\right)}\:\:\:-\:-\:[\:a^2\:-\:b^2\:=\:(\:a\:+\:b\:)\:(\:a\:-\:b\:)\:]$

$\\\\\\\displaystyle\implies\sf\:\sin\:x\:=\:\sqrt{\left(\:\dfrac{13\:+\:5}{13}\:\right)\:\left(\:\dfrac{13\:-\:5}{13}\:\right)}\\\\\\\implies\sf\:\sin\:x\:=\:\sqrt{\dfrac{18}{13}\:\times\:\dfrac{8}{13}}\\\\\\\implies\sf\:\sin\:x\:=\:\sqrt{\dfrac{18\:\times\:8}{13\:\times\:13}}\\\\\\\implies\sf\:\sin\:x\:=\:\sqrt{\dfrac{9\:\times\:2\:\times\:8}{13\:\times\:13}}\\\\\\\implies\sf\:\sin\:x\:=\:\sqrt{\dfrac{9\:\times\:16}{13\:\times\:13}}\\\\\\\implies\sf\:\sin\:x\:=\:\dfrac{3\:\times\:4}{13}\\\\\\\implies\boxed{\red{\sf\:\sin\:x\:=\:\dfrac{12}{13}}}$

Now, by solving the LHS of the given equation, we get,

$\displaystyle\sf\:LHS\:=\:\dfrac{\left(\:2\:\sin\:x\:-\:3\:\cos\:x\:\right)}{\left(\:4\:\sin\:x\:-\:9\:\cos\:x\:\right)}\\\\\\\implies\sf\:LHS\:=\:\dfrac{\left(\:2\:\times\:\dfrac{12}{13}\:-\:3\:\times\:\dfrac{5}{13}\:\right)}{\left(\:4\:\times\:\dfrac{12}{13}\:-\:9\:\times\:\dfrac{5}{13}\:\right)}\\\\\\\implies\sf\:LHS\:=\:\dfrac{\left(\:\dfrac{24}{13}\:-\:\dfrac{15}{13}\:\right)}{\left(\:\dfrac{48}{13}\:-\:\dfrac{45}{13}\:\right)}\\\\\\\implies\sf\:LHS\:=\:\dfrac{\dfrac{24\:-\:15}{13}}{\dfrac{48\:-\:45}{13}}\\\\\\\implies\sf\:LHS\:=\:\dfrac{\dfrac{9}{13}}{\dfrac{3}{13}}\\\\\\\implies\sf\:LHS\:=\:\dfrac{9}{\cancel{13}}\:\times\:\dfrac{\cancel{13}}{3}\\\\\\\implies\sf\:LHS\:=\:\cancel{\dfrac{9}{3}}\\\\\\\implies\boxed{\red{\sf\:LHS\:=\:3}}\\\\\\\implies\sf\:RHS\:=\:3\\\\\\\therefore\boxed{\red{\sf\:LHS\:=\:RHS}}$

Hence proved!

Answer 2

Answer:

3

Explanation:

Given,

$sec\theta = \frac{13}{5} \:$

____________________

We know the Trigonometric identity:

tan² theta = sec² theta -1

= (13/5)² - 1

= 169/25 - 1

= (169 - 25)/25

= 144/25

tan theta = √(12/5)² = 12/5

_______________________

$tan\theta = \frac{12}{5} \: \\ \\ </p><p>LHS</p><p>=</p><p>\frac{2sin\theta -3cos\theta}{4sin\theta - 9cos\theta} \:$

$Divide \: numerator \: and \: \\ denominator \: by \: </p><p>cos\theta \: </p><p> ,we \: get$

$= \frac{2tan\theta - 3}{4tan\theta - 9} \: \\ </p><p>= \\ \\ </p><p>\frac{2 \times \frac{12}{5}-3}{4 \times \frac{12}{5}-9} \: \\ \\ </p><p>=</p><p>\frac{\frac{24-15}{5}}{\frac{48-45}{5}} \\ \\ </p><p>= </p><p>\frac{24-15}{48-45}</p><p>= </p><p>\frac{9}{3} \\ \\ </p><p>= </p><p>3 \\ \\ </p><p>= RHS$

$</p><p>= \: </p><p>3 \\ </p><p>= RHS \:$

hope this helps you