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When a body is subjected to a direct tensile stress ($$\sigma $$) in one plane, then tangential or shear stress on an oblique section of the body inclined at an angle $$\theta $$ to the normal of the section is

When a body is subjected to a direct tensile stress ($$\sigma $$) in one plane, then tangential or shear stress on an oblique section of the body inclined at an angle $$\theta $$ to the normal of the section is Correct Answer $$\frac{\sigma }{2}\sin 2\theta $$

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$$\frac{{\cos 2\theta \cdot \cos 3\theta - \cos 2\theta \cdot \cos 7\theta + \cos \theta \cdot \cos 10\theta }}{{\sin 4\theta \cdot \sin 3\theta - \sin 2\theta \cdot \sin 5\theta + \sin 4\theta \cdot \sin 7\theta }}$$
When a body is subjected to a direct tensile stress ($$\sigma $$) in one plane, then normal stress on an oblique section of the body inclined at an angle $$\theta $$ to the normal of the section is
The value of $$\frac{{\sec \theta \left( {1 - \sin \theta } \right)\left( {\sin \theta + \cos \theta } \right)\left( {\sec \theta + \tan \theta } \right)}}{{\sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {1 + \cot \theta } \right)}}$$        is equal to:
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$$\frac{{1 + \cos \theta - {{\sin }^2}\theta }}{{\sin \theta \left( {1 + \cos \theta } \right)}} \times \frac{{\sqrt {{{\sec }^2}\theta + {\text{cose}}{{\text{c}}^2}\theta } }}{{\tan \theta + \cot \theta }},$$       0°
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