InterviewSolution
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InterviewSolution
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\(\frac{{{{\sec }^2}\theta- {{\cot }^2}\left( {90^\circ- \theta } \right)}}{{cose{c^2}67^\circ- {{\tan }^2}23^\circ }} + {\sin ^2}40^\circ+ {\sin ^2}50^\circ\)is equal to1). 02). 43). 24). 1 |
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Answer» TAN(90° - θ) = cotθ Sin(90° - θ) = cosθ Sin2θ + cos2θ = 1 Sec2θ – tan2θ = 1 cosec2θ – COT2θ = 1 Given, $(\frac{{{{\sec }^2}\theta- {{\cot }^2}\left( {90^\circ- \theta } \right)}}{{COSE{c^2}67^\circ- {{\tan }^2}23^\circ }} + {\sin ^2}40^\circ+ {\sin ^2}50^\circ)$ $(= \frac{{{{\sec }^2}\theta- {{\tan }^2}\theta }}{{cose{c^2}67^\circ- {{\tan }^2}(90^\circ- \;67^\circ )}} + {\sin ^2}40^\circ+ {\sin ^2}\left( {90^\circ- 40^\circ } \right))$ $(= \frac{1}{{cose{c^2}67^\circ- {{\cot }^2}67^\circ }} + \;{\sin ^2}40^\circ+ co{s^2}40^\circ)$ = 1 + 1 = 2 |
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