Question

please answer this question tomorrow is my exam

Answer 1

Given Equation : ( cos58° / sin32° ) + ( sin22° / cos68° ) - [ { ( cos38° cosec52° ) / ( tan18° tan35° tan60° ) } cos72° cot55° ]



Now,

= > { cos( 90 - 32 )° / sin32° } + { sin( 90 - 68 )° / cos68° } - [ { cos( 90 - 52 )° cosec52° } / { tan( 90 - 72 )° tan( 90 - 55 )° tan60° } ( cos72° cot55° ) ]


From the properties of trigonometric ratios, we know : -

sin( 90 - A ) = cosA
cos( 90 - A ) = sinA
tan( 90 - A ) = cotA


Then,

= > { sin32° / sin32° } + { cos 68° / cos68° } - [ { sin52° cosec52° } / { cot72° cot55° tan60° } ( cos72° cot55° ) ]


= > { 1 } + { 1 } - [ 1 / ( cot72° cot55° tan60° ) x ( cos72° cot55° ) ]


= > ( 1 + 1 ) - [ cos72° / ( cot72° tan60° ) ]


From the trigonometric table,
tan60° = √3


= > 2 - [ cos72° / cot72° x ( √3 ) ]


= > [ 2√3 - { cos72° / ( cos72° / sin72° ) } ] / ( √3 )


= > [ 2√3 - sin72° ] / √3




But, if the question is incorrect and correct Question is ( cos58° / sin32° ) + ( sin22° / cos68° ) - [ { ( cos38° cosec52° ) / ( tan18° tan35° tan60° ) } cot72° cot55° ]



Solution will be : -

= > { cos( 90 - 32 )° / sin32° } + { sin( 90 - 68 )° / cos68° } - [ { cos( 90 - 52 )° cosec52° } / { tan( 90 - 72 )° tan( 90 - 55 )° tan60° } ( cot72° cot55° ) ]


From the properties of trigonometric ratios, we know : -

sin( 90 - A ) = cosA
cos( 90 - A ) = sinA
tan( 90 - A ) = cotA


Then,

= > { sin32° / sin32° } + { cos 68° / cos68° } - [ { sin52° cosec52° } / { cot72° cot55° tan60° } ( cot72° cot55° ) ]


= > { 1 } + { 1 } - [ 1 / ( cot72° cot55° tan60° ) x ( cot72° cot55° ) ]


= > ( 1 + 1 ) - [ cot72° / ( cot72° tan60° ) ]


From the trigonometric table,
tan60° = √3


= > 2 - [ cot72° / cot72° x ( √3 ) ]


= > 2 - ( 1 / √3 )

= > $\dfrac{2\sqrt{3}-1}{\sqrt{3}}$
$\:$

Answer 2

(cos 58°/sin 32°)+(sin 22°/cos 68°)-(cos 38°cosec 52°/tan 18° tan 35° tan 60°)cos 72° cot 55°

Let us solve step by step
==
==
Let's first solve (cos 58°/ sin 32°)

→cos(90-32)/sin 32=>sin 32/sin 32=>1
==
==
Now (sin 22°/ cos 68°)

→sin(90-68)/cos 68=>cos 68/cos 68=>1
==
==
And now last part

→cos(90-52) cosec 52/tan 18° tan 35° tan 60° tan 72° tan 55°
→sin 52 cosec 52/ tan 18° tan 35° tan 60° tan 72° tan 55°
→sin 52 cosec 52/ tan 18 tan 72 tan 35 tan 55 tan 60
→cosec 52=1/sin 52
→1/tan(90-18)tan 72 tan (90-35)tan 55 tan 60°
1→/ cot 72 tan 72 cot 55 tan 55 tan 60

→Cot 72 tan 72 are cancelled
→Cot 55 tan 55 again cancelled

→1/1×1 tan 60
→1/tan 60
==
==
Combining each part together

→1+1-1/tan 60
→2-1/tan60
Rationalising 1/√3
1/√3×(√3/√3)=>√3/3
(2-√3)/3
(6-√3)/3

(6-√3)/3 is the Answer