Question

sin 90 degree minus theta cos 90 degree minus theta is equals to 10 theta upon 1 + 10 square theta​

Answer 1

$\textbf{To prove:}$

$sin(90^{\circ}-\theta)\,cos(90^{\circ}-\theta)=\dfrac{tan\,\theta}{1+\tan^2\theta}$

$\text{Consider,}$

$sin(90^{\circ}-\theta)\,cos(90^{\circ}-\theta)$

$=cos\,\theta\,sin\,\theta$

$\text{Multiply both numerator and denominator by cos\theta }$

$=cos^2\theta\,\dfrac{sin\,\theta}{cos\,\theta}$

$=cos^2\theta\,tan\,\theta$

$=\dfrac{tan\,\theta}{sec^2\theta}$

$=\dfrac{tan\,\theta}{1+tan^2\theta}$

$\therefore\bf\,sin(90^{\circ}-\theta)\,cos(90^{\circ}-\theta)=\dfrac{tan\,\theta}{1+\tan^2\theta}$

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Answer 2

Answer:

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