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The value of \(\frac{1}{{\left( {1 + {{\tan }^2}\theta } \right)}} + \frac{1}{{\left( {1 + {{\cot }^2}\theta } \right)}}\) is1). 1/42). 13). 24). 1/2 |
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Answer» Given expression is, $(\frac{1}{{\left( {1 + {{\TAN }^2}\theta } \right)}} + \frac{1}{{\left( {1 + {{\cot }^2}\theta } \right)}})$ Using the property: 1 + tan2θ = sec2θ and 1 + cot2θ = COSEC2θ $(= \frac{1}{{se{c^2}\theta }} + \frac{1}{{COSE{c^2}\theta }})$ Using the property: sec θ = 1/cosθ and cosec θ = 1/sin θ $(= \frac{1}{{\frac{1}{{co{s^2}\theta }}}} + \frac{1}{{\frac{1}{{SI{n^2}\theta }}}})$ = cos2θ + sin2θ ? sin2θ + cos2θ = 1 = 1 |
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