Question

Evaluate: 8√3cosec²30° sin60° cos60° cos²45°sin 45°tan30°cosec³45°
(class 10 CBSE SAMPLE PAPER 2017-18 MATHS)

Answer 1

SOLUTION:

8√3cosec²30° sin 60° cos 60° cos²45°sin 45°tan 30°cosec³45°

= 8√3(1/sin²30°). sin 60° cos 60° cos²45°sin 45°(sin 30°/cos 30°). 1/sin³45°


[ Cosec A= 1/sin A, tan A= sinA/cosA]

= 8√3(1/sin²30°)(sin 30°) sin 60° cos 60° cos²45°(1/cos 30°). sin 45°/sin³45°

= 8√3(1/sin 30°) sin 60° cos 60° cos²45°(1/cos 30°). 1/sin²45°

= 8√3(1/1/2) (√3/2)  (½) (1/√2)²(1/√3/2). 1/(1/√2)²


[ sin 30°= 1/2, sin 60° = √3/2, cos 60°=1/2,cos 45°= 1/√2, cos 30°=√3/2, sin 45°= 1/√2]


= 8√3 (2×√3×2×2) / (2×2×2×√3)

= 8×2×3×2×2 /8√3

= 24/√3

[On multiply numerator and denominator by √3]

= 24√3 / √3×√3= 24√3/3= 8√3


HOPE THIS WILL HELP YOU...

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

8√3cosec²30° × sin 60° × cos 60° × cos²45° × sin 45° × tan 30°× cosec³45°

Now,

We know that

$\boxed{\begin{minipage}{11 cm} Rule \\ \\ cosecA =\dfrac{1}{sinA} \\ \\ tanA=\dfrac{sinA}{cosA} \end{minipage}}$

$\tt{\rightarrow 8\sqrt{3}\dfrac{1}{sin^2 30}\times sin60\times cos60\times cos^2 45\times sin45\dfrac{sin30}{cos30}\times\dfrac{1}{sin^3 45}}$

Also,

$\tt{\rightarrow 8\sqrt{3}\times(\dfrac{1}{sin^2 30})(sin30)\times sin60\times sin60\times cos^2 45 (\dfrac{1}{cos30})\times\dfrac{sin45}{sin^3 45}}$

$\tt{\rightarrow 8\sqrt{3}(\dfrac{1}{sin30})\times sin60\times cos60\times cos^2 45(\dfrac{1}{cos30})\times\dfrac{1}{sin^2 45}}$

= 8√3(1/1/2) (√3/2) × (1/2) × (1/√2)² × (1/√3/2) × 1/(1/√2)²

$\tt{\rightarrow 8\sqrt{3}\times\dfrac{2\times\sqrt{3}\times 2\times 2}{2\times 2\times 2\times\sqrt{3}}}$

$\tt{\rightarrow\dfrac{8\times 2\times 3\times 2\times 2}{8\sqrt{3}}}$

$\tt{\rightarrow\dfrac{24}{\sqrt{3}}}$

$\tt{\rightarrow\dfrac{24\sqrt{3}}{\sqrt{3}\times\sqrt{3}}}$

$\tt{\rightarrow\dfrac{24\sqrt{3}}{3}}$

= 8√3