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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

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 Correct Answer $$\sigma {\cos ^2}\theta $$

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Evaluate the following:
$$\frac{{\cos 2\theta \cdot \cos 3\theta - \cos 2\theta \cdot \cos 7\theta + \cos \theta \cdot \cos 10\theta }}{{\sin 4\theta \cdot \sin 3\theta - \sin 2\theta \cdot \sin 5\theta + \sin 4\theta \cdot \sin 7\theta }}$$
When a body is subjected to a direct tensile stress ($$\sigma $$) in one plane, then tangential or shear 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:
If $$\frac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }} = \frac{1}{7},\,\theta ,$$     lies in first quadrant, then the value of $$\frac{{{\text{cosec}}\,\theta + {{\cot }^2}\theta }}{{{\text{cosec}}\,\theta - {{\cot }^2}\theta }}$$   is:
$$\frac{{{{\left( {1 + \sec \theta \,{\text{cosec}}\,\theta } \right)}^2}{{\left( {\sec \theta - \tan \theta } \right)}^2}\left( {1 + \sin \theta } \right)}}{{{{\left( {\sin \theta + \sec \theta } \right)}^2} + {{\left( {\cos \theta + {\text{cosec}}\,\theta } \right)}^2}}},$$        0°
The expression $$\frac{{{{\left( {1 - \sin \theta + \cos \theta } \right)}^2}\left( {1 - \cos \theta } \right){{\sec }^3}\theta \,{\text{cose}}{{\text{c}}^2}\theta }}{{\left( {\sec \theta - \tan \theta } \right)\left( {\tan \theta + \cot \theta } \right)}},$$        0°
If $$\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta - 3\cos \theta + 2}} = 1,\,\theta $$     lies in the first quadrant, then the value of $$\frac{{{{\tan }^2}\frac{\theta }{2} + {{\sin }^2}\frac{\theta }{2}}}{{\tan \theta + \sin \theta }}$$    is:
What is the value of [(cos 3θ + 2cos 5θ + cos 7θ)÷(cos θ + 2cos 3θ + cos 5θ)] + sin 2θ tan 3θ?
$$\frac{{1 + \cos \theta - {{\sin }^2}\theta }}{{\sin \theta \left( {1 + \cos \theta } \right)}} \times \frac{{\sqrt {{{\sec }^2}\theta + {\text{cose}}{{\text{c}}^2}\theta } }}{{\tan \theta + \cot \theta }},$$       0°
The expression $$\frac{{\left( {1 - 2{{\sin }^2}\theta {{\cos }^2}\theta } \right)\left( {\cot \theta + 1} \right)\cos \theta }}{{\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right)\left( {1 + \tan \theta } \right){\text{cosec}}\,\theta }} - 1,$$       0°