Question

$\Large{\underline{\underline{\bf{Question:-}}}}$
Find the value of $\sqrt{3}$ cosec 20° – sec 20°.​

Answer 1

☀️ Given that, we have to find the value of √3 cossec 20° - sec 20°.

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T H E R E F O R E,

As we know that,

$\bigstar \;\; \large{\underline{\boxed{ \bf cossec \theta = \dfrac{1}{sin \theta \; }}}}$

$\bigstar \;\; \large{\underline{\boxed{ \bf sec \theta = \dfrac{1}{cos \theta \; }}}}$

$\sf : \; \implies \dfrac{\sqrt{3}}{sin \; 20^{ \circ }} - \dfrac{1}{cos \; 20^{ \circ}}$

$\sf : \; \implies 2 \bigg( \dfrac{ \dfrac{\sqrt{3}}{2}}{sin \; 20^{ \circ }} - \dfrac{\dfrac{1}{2}}{cos \; 20^{ \circ}} \bigg)$

$\bigstar \;\; \large{\underline{\boxed{ \bf cos \; 30^{ \circ } = sin \; 60^{ \circ } =\dfrac{ \sqrt{3}}{2} \; }}}}$

$\sf : \; \implies 2 \bigg( \dfrac{cos\; 30^{\circ}}{sin \; 20^{ \circ }} - \dfrac{sin \; 30^{\circ}}{cos \; 20^{ \circ}} \bigg)$

$\sf : \; \implies 2 \bigg( \dfrac{cos\; 30^{\circ} \; cos\; 20^{\circ} - sin \; 30^{\circ} \; sin \; 20^{\circ}}{sin \; 20^{ \circ } \; cos \; 20^{\circ}} \bigg)$

$\bigstar \;\; \large{\underline{\boxed{ \bf (cos \; A \times cos \; B )- ( sin \; A \times sin \; B ) = cos( A + B )\; }}}}$

$\sf : \; \implies 2 \bigg( \dfrac{cos\; ( 50^{\circ}) }{sin \; 20^{ \circ } \; cos \; 20^{\circ}} \bigg)$

$\sf : \; \implies 2 \times 2 \bigg( \dfrac{cos\; ( 50^{\circ}) }{ 2 \times sin \; 20^{ \circ } \; cos \; 20^{\circ}} \bigg)$

$\bigstar \;\; \large{\underline{\boxed{ \bf 2 \times sin \; A \times cos \; A = sin ( 2A )\; }}}}$

$\sf : \; \implies 4 \bigg( \dfrac{cos\; ( 50^{\circ}) }{sin \; 40^{ \circ}} \bigg)$

$\sf : \; \implies 4 \bigg( \dfrac{cos\; ( 50^{\circ}) }{sin \; (90^{ \circ} - 50^{\circ})} \bigg)$

$\bigstar \;\; \large{\underline{\boxed{ \bf sin ( 90 - \theta ) = cos \theta\; }}}}$

$\sf : \; \implies 4 \bigg( \dfrac{cos\; ( 50^{\circ}) }{cos50^{\circ}} \bigg)$

$\sf : \; \implies 4 \times 1$

$\boxed{\bf{ \maltese \;\; \sqrt{3}\; cossec \; 20^{\circ} - sec\; 20^{\circ} \implies 4 }}$

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• Henceforth, the value is 4.

Add to knowledge :-

$\boxed{\begin{minipage}{7 cm}Fundamental Trigonometric Identities \\ \\ \sin^2\theta + \cos^2\theta=1 \\ \\1+\tan^2\theta = \sec^2\theta \\ \\1+\cot^2\theta = \text{cosec}^2 \, \theta \end{minipage}}$

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Answer 2

Answer: $√3 cosec 20° - sec 20°= √3/ sin 20° - 1/ cos 20°= √3cos 20° - sin 20°/ sin 20° cos 20°= 4 [√3/2 cos 20° - 1/2 sin 20° / 2 sin 20° cos 20°]= 4 [ sin 60° cos 20° - cos 60° sin 20°/ sin 40°] -- ( 2 sin A cos A = sin 2A)Using identity sin ( A - B ) = sin A cos B - cos A sin B= 4 sin (60° - 20°) / sin 40°= 4 sin 40°/ sin 40°= 4$

√3 cosec 20° - sec 20°

= √3/ sin 20° - 1/ cos 20°

= √3cos 20° - sin 20°/ sin 20° cos 20°

= 4 [√3/2 cos 20° - 1/2 sin 20° / 2 sin 20° cos 20°]

= 4 [ sin 60° cos 20° - cos 60° sin 20°/ sin 40°] -- ( 2 sin A cos A = sin 2A)

Using identity sin ( A - B ) = sin A cos B - cos A sin B

= 4 sin (60° - 20°) / sin 40°

= 4 sin 40°/ sin 40°

= 4

Step-by-step explanation:

its my answer