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In a right triangle ABC, right angled at C, if∠B = 60° and AB = 15 units. Therefore,

And,

Given that1+cot2α/(1+cosecα)=cosecα

=LHS =RHS Hence proved

First constructing the triangle to get the diagram,

Drawing BC a horizontal line,

But actually nothing is mentioned about BC's length.

But we only know that∠BAC = 60˚

And knowing that it is a right-angled triangle we can say that one angle is 90˚ and another is180° - 90° - 60° = 30°

So, two possible cases arise,

if∠B = 90˚ and∠C = 30˚, (Condition 1)

then BC > 8m

if∠B =30˚ and∠C= 90˚, (Condition 2)

then BC < 8m

Actual understanding of this requires knowing equation of circles and lines and equating them which results in two variables and because of the two predefined conditions which are independent we get two conditions.

Solving condition 1,

Here, AB = 8m and∠CBA= 90˚ and∠BAC= 60˚ and∠ACB=30˚.

Angle of elevation of A from C will actually be equal to the∠BCA

where BC acts as the reference line from which the elevated angle is measured.

Therefore, angle of elevation of A from C =∠BCA = 60˚

A horizontal line parallel to BC is drawn from A such that it acts like a reference line for calculating the angle of depression of C from A.

Angle of depression of C from A =∠CAD

Now,∠CAD =∠ACB as they are alternate angles where AD and BC acts as the parallel lines.

Therefore,∠CAD =∠ACB = 30˚

Therefore, the angle of depression = 60˚

Now distance of B from C i.e; BC can be calculated from trigonometric ratios.

Using,

BC =8√3

Now,

Solving for Condition 2,

Here, angle BAC = 60˚

∠ABC = 30˚

∠ACB = 90˚.

Also, here, BC < AB.

Now, using the above formulas and concepts for these cases we will again get the corresponding answers.

Angle of elevation of A at C = 30˚

Angle of depression of C at A= 60˚

Length of BC =4√3

Given thatcotθ- tanθ=(2cos2θ-1)/(sinθcosθ).

In ∆ABD,

Using Pythagoras theorem, we get

= 6 cm

Again,

In ∆ABC,

Using Pythagoras therem, we get

Now,

=LHS = RHS Hence proved