Question

sin30°+tan45°-cosec60°/sec30 +cos60 +cot 45 evaluate​

Answer 1

Answer:

answer is given for your problem

Answer 2

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$\sf\dfrac{sin30°+tan45°-cosec60°}{sec30 +cos60 +cot 45}$

evaluate

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Solution}}}}}}}}$

$\sf\dfrac{sin30°+tan45°-cosec60}{sec30 +cos60 +cot 45}$

• $\sf sin 30 = \dfrac{1}{2}$

• $\sf tan 45 = 1$

• $\sf cosec 60 = \dfrac{2}{\sqrt3}$

• $\sf sec 30 = \dfrac{2}{\sqrt3}$

• $\sf cos 60 = \dfrac{1}{2}$

• $\sf cot 45 = 1$

$\implies\sf\dfrac{\dfrac{1}{2} + 1 - \dfrac{2}{\sqrt3}} {\dfrac{2}{\sqrt3} + \dfrac{1}{2} + 1}$

$\implies\sf\dfrac{\dfrac{\sqrt3 + 2\sqrt3 - 4 }{2\sqrt3}}{\dfrac{4 + \sqrt3 + 2\sqrt3}{2\sqrt3}}$

$\implies\sf{\dfrac{\sqrt3 + 2\sqrt3 - 4 }{\cancel{2\sqrt3}}}\times{\dfrac{\cancel{2\sqrt3}}{4 + \sqrt3 + 2\sqrt3}}$

$\implies\sf\dfrac{\sqrt3 + 2\sqrt3 - 4 }{4+\sqrt3+2\sqrt3}$

$\implies\sf\dfrac{ 3\sqrt3 - 4 }{4+3\sqrt3}$

• rationalize the denominator

$\implies\sf\dfrac{ 3\sqrt3 - 4 }{4+3\sqrt3}\times\dfrac{ 3\sqrt3 - 4 }{3\sqrt3-4}$

• $(a+b) (a-b) = a^2 - b^2$

$\implies \dfrac{(3\sqrt3-4)^2}{(3\sqrt3)^2 + (4)^2}$

• $(a+b)^2 = a^2 + b^2 + 2ab$

$\implies\sf\dfrac{27+16-24\sqrt3}{27-16}$

$\implies\sf\dfrac{43- 24\sqrt3}{11}$

${\bold{\blue{\boxed{\bf{\dfrac{43- 24\sqrt3}{11}}}}}}$

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