Question

without using trigonometry table evaluate the following​

Answer 1

$\colorbox{skyblue} {Ans.13}$

$\frac{\sin15 ° \cos75° + \cos15° \sin75° }{ \tan5° \tan30 ° \tan35° \tan55° \tan85 ° }$

$= \frac{ \sin15° \cos(90° - 15°) + \cos15° \sin(90° - 15°) }{ \tan5° \frac{1}{ \sqrt{3} }. \tan35 ° \tan(90° - 35°) . \tan(90° - 5) }$

$\left(\cos(90° -Ø ) = \sinØ; \sin(90° -Ø ) = \cosØ; \tan(90° -Ø ) = \cotØ; \tanØ = \frac{1}{ \cotØ}\right)$

$= \frac{ \sin15 °. \sin15° + \cot15°. \cos15 ° }{ \frac{1}{ \cot5°} . \frac{1}{ \sqrt{3} } . \frac{1}{ \cot35° }. \cot35°. \cot5 °}$

$\left\{ { \sin}^{2Ø} + { \cos }^{2Ø} = 1\right\}$

$⇒ \frac{ { \sin }^{2}15° + { \cos}^{2} 15°}{ \frac{1}{ \sqrt{3} } } = 1 \times \frac{ \sqrt{3} }{1} = \sqrt{3}$

$\colorbox{skyblue} {Ans.14}$

$\frac{3 \cos55° }{7 \sin35° } - \frac{4( \cos(70°. \csc20° ) }{7( \tan5° \tan25°. \tan45°. \tan65°. \tan85° ) }$

$= \frac{3 \sin(90° - 35°) }{7 \sin35 °} - \frac{4 \cos70°. \sec(90 ° - 20°)}{7( \tan5°. \tan25°.1 \times \cot(90° - 65°). \cot(90° - 85°) }$

$= \frac{3 \sin35 °}{7 \sin35° } - \frac{4. \frac{1}{ \sec70° } . \sec70° }{7 \times \frac{1}{ \cot5 °} . \frac{1}{ \cot25° } \times 1 \times \cot25 °. \cot5 °}$

$⇒ \frac{3}{7} - \frac{4 \times 1}{7 \times 1} = \frac{3}{7} - \frac{4}{7} = \frac{ - 1}{7}$

$\\ \\ \\ \colorbox{lightgreen} {\red★ANSWER ᵇʸɴᴀᴡᴀʙ⌨}$

Answer 2

Answer:

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