Question

cos ( 90+a) sec ( -a) tan ( 180-a) / sec ( 360-a) sin ( 180-a) cot ( 90-a ) =1

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Answer 1

$\textbf{Given:}$

$\displaystyle\frac{cos(90^{\circ}+A)\,sec(-A)\,tan(180^{\circ}-A)}{sec(360^{\circ}-A)\,sin(180^{\circ}-A)\,cot(90^{\circ}-A)}$

$\textbf{To prove:}$

$\displaystyle\frac{cos(90^{\circ}+A)\,sec(-A)\,tan(180^{\circ}-A)}{sec(360^{\circ}-A)\,sin(180^{\circ}-A)\,cot(90^{\circ}-A)}=1$

$\boxed{\bf\,cos(90^{\circ}+A)=-sinA}$

$\boxed{\bf\,sec(-A)=secA}$

$\boxed{\bf\,tan(180^{\circ}-A)=-tanA}$

$\boxed{\bf\,sec(360^{\circ}-A)=secA}$

$\boxed{\bf\,sin(180^{\circ}-A)=sinA}$

$\boxed{\bf\,cot(90^{\circ}-A)=tanA}$

$\text{Now}$

$\displaystyle\frac{cos(90^{\circ}+A)\,sec(-A)\,tan(180^{\circ}-A)}{sec(360^{\circ}-A)\,sin(180^{\circ}-A)\,cot(90^{\circ}-A)}$

$=\displaystyle\frac{(-sinA)\,secA\,(-tanA)}{secA\,sinA\,tanA}$

$=\displaystyle\frac{sinA\,secA\,tanA}{secA\,sinA\,tanA}$

$=1$

$\therefore\bf\displaystyle\frac{cos(90^{\circ}+A)\,sec(-A)\,tan(180^{\circ}-A)}{sec(360^{\circ}-A)\,sin(180^{\circ}-A)\,cot(90^{\circ}-A)}=1$

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