Question

cos 23 + sin 23 upon cos 23 minus sin 23​

Answer 1

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

$\mathsf{\dfrac{cos\,23^\circ+sin\,23^\circ}{cos\,23^\circ-sin\,23^\circ}}$

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

$\textsf{The value of}$

$\mathsf{\dfrac{cos\,23^\circ+sin\,23^\circ}{cos\,23^\circ-sin\,23^\circ}}$

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

$\underline{\textbf{Formula used:}}$

$\boxed{\mathsf{tan(A+B)=\dfrac{tanA+tanB}{1-tanA\,tanB}}}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{cos\,23^\circ+sin\,23^\circ}{cos\,23^\circ-sin\,23^\circ}}$

$\textsf{This can be written as,}$

$\mathsf{=\dfrac{cos\,23^\circ\left(1+\dfrac{sin\,23^\circ}{cos\,23^\circ}\right)}{cos\,23^\circ\left(1-\dfrac{sin\,23^\circ}{cos\,23^\circ}\right)}}$

$\mathsf{=\dfrac{1+tan\,23^\circ}{1-tan\,23^\circ}}$

$\mathsf{=\dfrac{tan\,45^\circ+tan\,23^\circ}{1-tan\,45^\circ\,tan\,23^\circ}}$

$\textsf{Using the above formula, we get}$

$\mathsf{=tan(45^\circ+23^\circ)}$

$\mathsf{=tan\,68^\circ}$

$\implies\boxed{\mathsf{\dfrac{cos\,23^\circ+sin\,23^\circ}{cos\,23^\circ-sin\,23^\circ}=tan\,68^\circ}}$