Question

SOLVE :-
${ \implies{ \red{ \dfrac{ \cos(45 \degree) }{ \sec(30 \degree) +\csc(30 \degree) } }}}$

PROVE THAT :-
${ \implies{ \red{ \dfrac{ \tan( \theta) }{1 - \cot( \theta) } + \dfrac{ \cot( \theta) }{1 - \tan( \theta) } = 1 + \sec( \theta) \times \csc( \theta) }}}$
​

Answer 1

$\huge\tt\underline\blue{thanks}$

$\huge \underline \frak \pink {★彡 refer attachment彡★}$

$\huge\tt\underline\blue{hope -its -helpful}$

$\huge \underline \frak \pink {★彡 Gavya彡★}$

Answer 2

Question 1 :

$\sf{ \dashrightarrow{ \red{ \dfrac{ \cos(45^{ \circ}) }{ \sec(30^{ \circ} ) + \csc(30 ^{ \circ}) } }}}$

Solution :

$\sf{ \longrightarrow \quad{ { \dfrac{ \cos(45^{ \circ}) }{ \sec(30^{ \circ} ) + \csc(30 ^{ \circ}) } }}}$

$\sf \longrightarrow \quad \dfrac{ \dfrac{1}{ \sqrt{2} } }{ \dfrac{2}{ \sqrt{3}} + 2}$

$\sf \longrightarrow \quad \dfrac{ \dfrac{1}{ \sqrt{2} } }{ \dfrac{2 + 2 \sqrt{3} }{\sqrt{3} } }$

$\sf \longrightarrow \quad \dfrac{1}{ \sqrt{2}} \times \dfrac{ \sqrt{3} }{2 + 2 \sqrt{3} }$

$\sf \longrightarrow \quad \dfrac{ \sqrt{3} }{2 \sqrt{2} + 2 \sqrt{6} }$

Now rationalize the denominator.

$\sf \longrightarrow \quad \dfrac{ \sqrt{3} \big(2 \sqrt{2} - 2 \sqrt{6} \big)}{ \big(2 \sqrt{2} + 2 \sqrt{6} \big) \big(2 \sqrt{2} - 2 \sqrt{6} \big)}$

Using Identity :-

$\sf \dashrightarrow \blue{(a - b)(a + b) = {a}^{2} - {b}^{2}}$

$\sf \longrightarrow \quad \dfrac{ \sqrt{3} \big(2 \sqrt{2} - 2 \sqrt{6} \big)}{ \big(2 \sqrt{2}\big)^2 - \big(2 \sqrt{6} \big)^{2}}$

$\sf \longrightarrow \quad \dfrac{ \sqrt{3} \big(2 \sqrt{2} - 2 \sqrt{6} \big)}{ \big(2 \sqrt{2}\big)\big(2 \sqrt{2}\big) - \big(2 \sqrt{6} \big)\big(2 \sqrt{6} \big)}$

$\sf \longrightarrow \quad \dfrac{2 \sqrt{6} - 2 \sqrt{18} }{8 - 24}$

$\sf \longrightarrow \quad \dfrac{2 \sqrt{6} - 2 \sqrt{18} }{ - 16}$

$\sf \longrightarrow \quad \dfrac{2 \big(\sqrt{6} - \sqrt{2 \times 3 \times 3} \big) }{ - 16}$

$\sf \longrightarrow \quad \dfrac{2 \big(\sqrt{6} - 3\sqrt{2} \big) }{ - 16}$

$\sf \longrightarrow \quad \dfrac{ \not2 \big(\sqrt{6} - 3\sqrt{2} \big) }{ - \not16}$

$\sf \longrightarrow \quad \dfrac{ \sqrt{6} - 3\sqrt{2} }{ - 8}$

$\sf \longrightarrow \quad \dfrac{ - \sqrt{6} + 3\sqrt{2} }{8}$

$\boxed{ \sf \pmb{ \gray {Answer \dashrightarrow \dfrac{ - 6 + 3 \sqrt{2} }{8} }}} \quad\red \bigstar$

Question 2 :

To prove :-

$\sf {\dashrightarrow{\red{ \dfrac{ \tan \theta}{1 - \cot \theta } + \dfrac{ \cot \theta }{1 - \tan\theta} = 1 + \sec \theta \times \csc\theta}}}$

Solution :

$\sf \longrightarrow \quad \dfrac{ \tan\theta }{1 - \cot \theta } + \dfrac{ \cot \theta }{1 - \tan \theta} = 1 + \sec \theta \times \csc \theta$

Taking LHS :

$\sf \longrightarrow \quad \dfrac{ \tan\theta }{1 - \cot \theta } + \dfrac{ \cot \theta }{1 - \tan \theta}$

$\sf \longrightarrow \quad \dfrac{ \dfrac{ \sin \theta }{ \cos \theta } }{1 - \cot \theta } + \dfrac{ \dfrac{ \cos \theta}{ \sin\theta } }{1 - \tan \theta}$

$\sf \longrightarrow \quad \dfrac{ \dfrac{ \sin \theta }{ \cos \theta } }{1 - \dfrac{ \cos \theta}{ \sin \theta} } + \dfrac{ \dfrac{ \cos \theta}{ \sin\theta } }{1 - \dfrac{ \sin \theta}{ \cos \theta} }$

$\sf \longrightarrow \quad \dfrac{ \dfrac{ \sin \theta }{ \cos \theta } }{\dfrac{ \sin \theta - \cos \theta}{ \sin \theta} } + \dfrac{ \dfrac{ \cos \theta}{ \sin\theta } }{ \dfrac{ \cos \theta - \sin \theta}{ \cos \theta} }$

$\sf \longrightarrow \quad \dfrac{ \sin \theta}{ \cos\theta} \times \dfrac{ \sin\theta}{ \sin\theta - \cos\theta} + \dfrac{ \cos\theta}{ \sin\theta} \times \dfrac{ \cos\theta}{ \cos\theta - \sin \theta}$

$\sf \longrightarrow \quad \dfrac{ \sin^2\theta}{ \cos \theta( \sin\theta - \cos\theta)} - \dfrac{ \cos^2\theta}{ \sin \theta(\sin\theta - \cos \theta)}$

Taking (sin θ - cos θ) common.

$\sf \longrightarrow \quad \dfrac{1}{( \sin \theta - \cos \theta)} \Bigg[ \dfrac{ \sin^{2} \theta }{ \cos \theta} - \dfrac{ { \cos ^{2}\theta } }{ \sin \theta} \Bigg]$

$\sf \longrightarrow \quad \dfrac{1}{( \sin \theta - \cos \theta)} \Bigg[ \dfrac{ \sin^{3} \theta - \cos^{3} \theta }{ \sin \theta \cos \theta} \Bigg]$

Using Identity :-

$\sf\dashrightarrow \quad \blue{{a}^{3} - {b}^{3} =(a - b)( {a}^{2} + {b}^{2} + ab)}$

$\sf \longrightarrow \quad \dfrac{1}{( \sin \theta - \cos \theta)} \Bigg[ \dfrac{ (\sin \theta - \cos \theta)( { \sin}^{2} \theta + { \cos}^{2} \theta + \sin \theta. \cos \theta }{ \sin \theta \cos \theta} \Bigg]$

$\sf \longrightarrow \quad \dfrac{1}{ \cancel{( \sin \theta - \cos \theta)}} \Bigg[ \dfrac{ \cancel{ (\sin \theta - \cos \theta)}( { \sin}^{2} \theta + { \cos}^{2} \theta + \sin \theta. \cos \theta }{ \sin \theta \cos \theta} \Bigg]$

$\sf \longrightarrow \quad \dfrac{ { \sin}^{2} \theta + { \cos}^{2} \theta + \sin \theta. \cos \theta }{ \sin \theta \cos \theta}$

$\sf \longrightarrow \quad \dfrac{1 + \sin \theta. \cos \theta }{ \sin \theta \cos \theta}$

$\sf \longrightarrow \quad \dfrac{1}{ \sin \theta \cos \theta} + \cancel\dfrac{ \sin \theta \cos \theta }{ \sin \theta \cos \theta}$

$\sf \longrightarrow \quad \sec \theta \csc \theta + 1$

$\sf LHS \longrightarrow \quad 1 + \sec \theta \csc \theta$

LHS = RHS

Hence, proved.