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Sia ? 6 years, 4 months ago
(1 + cotA + tanA) (sinA - cosA)
= sinA - cosA + cotA sinA - cotA cosA + sinA tanA - tanA cosA
{tex}= \sin A - \cos A + \frac{{\cos A}}{{\sin A}} \times \sin A - \cot A\cos A + \sin A\;\tan A - \frac{{\sin A}}{{\cos A}} \times \cos A{/tex}
= sinA - cosA + cosA - cotA cosA + sinA tanA - sinA
= sinA tanA - cotA cosA........(1)
Now taking ;
{tex}\quad \frac{{\sec A}}{{\cos e{c^2}A}} - \frac{{\cos ecA}}{{{{\sec }^2}A}}{/tex}
{tex} = \frac{{\frac{1}{{\cos A}}}}{{\frac{1}{{{{\sin }^2}A}}}} - \frac{{\frac{1}{{\sin A}}}}{{\frac{1}{{{{\cos }^2}A}}}}{/tex}
{tex} = \frac{{{{\sin }^2}A}}{{\cos A}} - \frac{{{{\cos }^2}A}}{{\sin A}}{/tex}
{tex} = \sin A \times \frac{{\sin A}}{{\cos A}} - \cos A \times \frac{{\cos A}}{{\sin A}}{/tex}
{tex} = \sin A \times \tan A - \cos A \times \cot A{/tex}.......(2)
From (1) & (2),
(1 + cotA + tanA) (sinA - cosA) = {tex}\frac { \sec A } { cosec ^ { 2 } A } - \frac { cosec A } { \sec ^ { 2 } A }{/tex} = sinA.tanA - cosA.cotA
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Ram Kushwah 6 years, 9 months ago
{tex}\begin{array}{l}\sin(45+\mathrm\theta)-\cos(45-\mathrm\theta)\\=\sin(45+\mathrm\theta)-\cos\lbrack90-(45+\mathrm\theta)\rbrack\\=\sin(45+\mathrm\theta)-\sin(45+\mathrm\theta)\\=0\end{array}{/tex}
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Sia ? 6 years, 4 months ago
Given, Radius = r
{tex} \triangle{/tex}AOB is a right triangle.
In the right {tex} \triangle{/tex}OAB,
{tex} \tan \theta = \frac { \mathrm { AB } } { \mathrm { OA } }{/tex}
{tex} \Rightarrow A B = O A \times \tan \theta{/tex}
{tex} = r \tan \theta{/tex}
Now, area of {tex} \triangle{/tex}OAB ={tex}\frac12 b \times h{/tex}={tex} \frac { 1 } { 2 } \mathrm { OA } \times \mathrm { AB }{/tex}
{tex} = \frac { 1 } { 2 } \times r \times r \tan \theta{/tex}
{tex} = \frac { 1 } { 2 } r ^ { 2 } \tan \theta{/tex}
and area of sector OAC ={tex} \pi r ^ { 2 } \times \frac { \theta } { 360 ^ { \circ } }{/tex}
length of arc AC = {tex} 2 \pi r \times \frac { \theta } { 360 ^ { \circ } }{/tex}
{tex} = \frac { 2 \pi r \theta } { 360 ^ { \circ } } = \frac { \pi r \theta } { 180 ^ { \circ } }{/tex}
{tex} \therefore{/tex}Perimeter of the shaded region
= Arc AC + AB + BC
{tex} = \frac { \pi r\theta } { 180 } + r \tan \theta + ( \mathrm { OB } - \mathrm { OC } ){/tex}
{tex} = \frac { \pi r\theta } { 180 } + r \tan \theta + ( r \sec \theta - r ){/tex}
{tex} = r \left( \frac { \pi \theta } { 180 } + \tan \theta + \sec \theta - 1 \right){/tex}
{tex} = r \left( \tan \theta + \sec \theta + \frac { \pi \theta } { 180 } - 1 \right){/tex}
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Devanshu Kumawat 6 years, 9 months ago
0Thank You