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JEE MATHEMATICS MCQ: TRANSFORMATIONS

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The least value of cosec2 x+25  sec⁡2 x\mathrm{cose}{\mathrm{c}^2}\,x + 25\,\,{\sec ^2}\,x is
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Correct option: D
38
Option D (38) is correct because the minimum value of the expression is obtained by completing the square and finding the minimum value of the squared term.

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If 1+cos⁡  56∘+cos⁡  58∘−cos⁡  66∘=λcos⁡  28∘  cos⁡  29∘  sin⁡  33∘1 + \cos \,\,56^\circ + \cos \,\,58^\circ - \cos \,\,66^\circ = \lambda \cos \,\,28^\circ \,\,\cos \,\,29^\circ \,\,\sin \,\,33^\circ , then λ=\lambda = The value of
cos⁡  π11+cos⁡  3 π11+cos⁡  5 π11+cos⁡  7π11+cos⁡  9 π11\cos \,\,\frac{\pi }{{11}} + \cos \,\,\frac{{3\,\pi }}{{11}} + \cos \,\,\frac{{5\,\pi }}{{11}} + \cos \,\,\frac{{7\pi }}{{11}} + \cos \,\,\frac{{9\,\pi }}{{11}}, is
If   cos⁡  x+cos⁡  y+cos⁡  α=0   and sin⁡x+sin⁡  y+sin⁡  α=0\;\cos \,\,x + \cos \,\,y + \cos \,\,\alpha = 0\;\,\mathrm{and}\,\sin x + \sin \,\,y + \sin \,\,\alpha = 0 then cot⁡(x+y2)  \cot \left( {\frac{{x + y}}{2}} \right)\; is equal to The period of the function f(x)=sin⁡(2x+36  π),f\left( x \right) = \sin \left( {\frac{{2x + 3}}{{6\;\pi }}} \right), is If 0<x,y<π0 < x,\mathrm{ }y < \pi and cos⁡x+cos⁡y−cos⁡(x+y)=32\cos x + \cos y - \cos (x + y) = \frac{3}{2}, then sin⁡x+cos⁡y\sin x + \cos y is equal to The value of   sin  36∘  sin⁡  72∘  sin⁡  108∘  sin⁡  144∘  \mathrm{\;sin}\,\,36^\circ \,\,\sin \,\,72^\circ \,\,\sin \,\,108^\circ \,\,\sin \,\,144^\circ \; is equal to
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