Question

sin30° + tan45° - cosec60° / sec30° + cos60° + cot45° evaluate. Plea

se solve this for 100 points.

Answer 1

Given :

• An equation, $\bf \dfrac{sin {30}^{ \circ} + tan {45}^{ \circ} - cosec {60}^{ \circ} }{sec {30}^{ \circ} + cos {60}^{ \circ} + cot {45}^{ \circ}}$

What to do ?

• To evaluate the given equation, $\bf \dfrac{sin {30}^{ \circ} + tan {45}^{ \circ} - cosec {60}^{ \circ} }{sec {30}^{ \circ} + cos {60}^{ \circ} + cot {45}^{ \circ}}$

Solution :

Given equation,

$\bf \implies\dfrac{sin {30}^{ \circ} + tan {45}^{ \circ} - cosec {60}^{ \circ} }{sec {30}^{ \circ} + cos {60}^{ \circ} + cot {45}^{ \circ} }$

We know that,

$\bf \bullet \: \: \: sin {30}^{ \circ} = \dfrac{1}{2} \\ \\ \bf \bullet \: \: \: tan {45}^{ \circ} = 1 \\ \\ \bf \bullet \: \: \: cosec {60}^{ \circ} = \dfrac{2}{ \sqrt{3} } \\ \\ \bf \bullet \: \: \: sec {30}^{ \circ} = \dfrac{2}{ \sqrt{3} } \\ \\ \bf \bullet \: \: \: cos {60}^{ \circ} = \dfrac{1}{2} \\ \\ \bf \bullet \: \: \: cot {45}^{ \circ} = 1$

Substitute all the values in the given equation.

$\bf \implies \dfrac{ \bigg(\dfrac{1}{2} \bigg)+( 1) - \bigg(\dfrac{2}{ \sqrt{3} } \bigg)}{ \bigg( \dfrac{2}{ \sqrt{3} } \bigg) + \bigg( \dfrac{1}{2} \bigg) + 1} \\ \\ \\ \bf \implies \dfrac{ \dfrac{1}{2} + 1 - \dfrac{2}{ \sqrt{3} } }{ \dfrac{2}{ \sqrt{3} }+ \dfrac{1}{2} + 1 } \\ \\ \\ \bf \implies \dfrac{ \dfrac{1 + 2}{2} - \dfrac{2}{ \sqrt{3}}}{ \dfrac{2}{ \sqrt{3}} + \dfrac{1 + 2}{2} }$

$\bf \implies \dfrac{ \dfrac{3}{2} - \dfrac{2}{ \sqrt{3}}}{ \dfrac{2}{ \sqrt{3}} + \dfrac{3}{2} } \\ \\ \\ \bf \implies \dfrac{ \bigg( \dfrac{3 \sqrt{3} - 4}{2 \sqrt{3} } \bigg)}{ \bigg( \dfrac{4 + 3 \sqrt{3}}{2 \sqrt{3} } \bigg)} \\ \\ \\ \bf \implies \dfrac{(3 \sqrt{3}- 4)}{(4 + 3 \sqrt{3})} \times \dfrac{ \not2\sqrt{3} }{ \not2\sqrt{3}}$

$\bf \implies \dfrac{(3 \sqrt{3} - 4)}{(4 + 3 \sqrt{3})} \\ \\ \\ \bf \implies \dfrac{(3 \sqrt{3} - 4)}{(3 \sqrt{3} + 4)} \times \dfrac{(3 \sqrt{3} - 4)}{(3 \sqrt{3} - 4)} \\ \\ \\ \bf \implies \dfrac{ {(3 \sqrt{3} - 4)}^{2} }{(3 \sqrt{3} + 4)(3 \sqrt{3} - 4)}$

Using two identities in the numerator and the denominator,

• (a - b)² = a² - 2ab + b²
• (a + b)(a - b) = a² - b²

$\bf \implies \dfrac{ {(3 \sqrt{3})}^{2} - 2(3 \sqrt{3})(4) + {(4)}^{2}}{ {(3 \sqrt{3})}^{2} - {(4)}^{2} } \\ \\ \\ \bf \implies \dfrac{9 \times 3 - 24 \sqrt{3} + 16}{9 \times 3 - 16} \\ \\ \\ \bf \implies \dfrac{27 - 24 \sqrt{3} + 16}{27 - 16} \\ \\ \\ \bf \implies \dfrac{27 + 16 - 24 \sqrt{3}}{11} \\ \\ \\ \bf \implies \dfrac{43 - 24 \sqrt{3} }{11}$

Hence,

$\bf \dfrac{sin {30}^{ \circ} + tan {45}^{ \circ} - cosec {60}^{ \circ} }{sec {30}^{ \circ} + cos {60}^{ \circ} + cot {45}^{ \circ}} = \dfrac{43 - 24 \sqrt{3} }{11}$

Answer 2

ɢɪᴠᴇɴ:-

• $\tt\frac{(sin30 + tan45 - cosec60)}{(sec30 + cos60 + cot45)}$

ᴛᴏ ᴅᴏ:-

• Evaluate

ꜱᴏʟᴜᴛɪᴏɴ:-

Substituting the values in the given equation,

$\begin{lgathered}\begin{lgathered}\\\end{lgathered}\end{lgathered}$

$:\implies\tt\frac{ \frac{1}{2}+1- \frac{2}{ \sqrt{3}} }{ \frac{2}{ \sqrt{3}}+ \frac{1}{2}+1}$

$\begin{lgathered}\begin{lgathered}\\\end{lgathered}\end{lgathered}$

$:\implies\tt\frac{ \sqrt{3}+2 \sqrt{3} - 4 }{4 + \sqrt{3} + 2 \sqrt{3} }$

$\begin{lgathered}\begin{lgathered}\\\end{lgathered}\end{lgathered}$

$:\implies\tt\frac{ 3 \sqrt{3} - 4 }{4 + 3 \sqrt{3} }$

Rationalising the denominator,

$\begin{lgathered}\begin{lgathered}\\\end{lgathered}\end{lgathered}$

$:\implies\tt\frac{ 3 \sqrt{3} - 4 }{4 + 3 \sqrt{3} }\times \frac{ 3 \sqrt{3} - 4 }{3 \sqrt{3} - 4 }$

$\begin{lgathered}\begin{lgathered}\\\end{lgathered}\end{lgathered}$

$:\implies\tt\frac{(3 \sqrt{3} - 4 )^2}{ (3 \sqrt{3} )^2-(4)^2 }$

$\begin{lgathered}\begin{lgathered}\\\end{lgathered}\end{lgathered}$

$:\implies\tt\frac{27+16-24 \sqrt{3} }{27-16}$

$\begin{lgathered}\begin{lgathered}\\\end{lgathered}\end{lgathered}$

$:\implies\bf\frac{43-24 \sqrt{3} }{11}$

$\begin{lgathered}\begin{lgathered}\\\end{lgathered}\end{lgathered}$

━━━━━━━━━━━━━━━━━━━━━━

$\sf \large \pink{Have\;a\;Look!!!}$

$\begin{lgathered}\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}\end{lgathered}$