Question

If 4 tan theta =3;then find value of (4sin theta -cos theta /4sin theta +cos theta )
Please solve this

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{4\;tan\theta=3}$

$\underline{\textbf{To find:}}$

$\textsf{The value of}$

$\mathsf{\dfrac{4\;sin\theta-cos\theta}{4\;sin\theta-cos\theta}}$

$\underline{\textbf{Solution:}}$

$\mathsf{4\;tan\theta=3}$

$\implies\mathsf{tan\theta=\dfrac{3}{4}}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{4\;sin\theta-cos\theta}{4\;sin\theta-cos\theta}}$

$\mathsf{By\;taking\;cos\theta\;as\;common\;from}$

$\textsf{both numerator and denominator}$

$\mathsf{=\dfrac{cos\theta\left(4\dfrac{sin\theta}{cos\theta}-1\right)}{cos\theta\left(4\dfrac{sin\theta}{cos\theta}+1\right)}}$

$\mathsf{=\dfrac{4\dfrac{sin\theta}{cos\theta}-1}{4\dfrac{sin\theta}{cos\theta}+1}}$

$\mathsf{=\dfrac{4\,tan\theta-1}{4\,tan\theta+1}}$

$\mathsf{=\dfrac{4\left(\dfrac{3}{4}\right)-1}{4\left(\dfrac{3}{4}\right)+1}}$

$\mathsf{=\dfrac{3-1}{3+1}}$

$\mathsf{=\dfrac{2}{4}}$

$\mathsf{=\dfrac{1}{2}}$

$\implies\boxed{\mathsf{\dfrac{4\;sin\theta-cos\theta}{4\;sin\theta-cos\theta}=\dfrac{1}{2}}}$

$\underline{\textbf{Find more:}}$

If 5cot A= 8, then the value of sinA secA is​

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