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Question: The value of the expression \( \cos ec(75^\circ + \theta ) - \sec (15^\circ - \theta ) - \tan (55^\c...

The value of the expression cos⁡ec(75∘+θ)−sec⁡(15∘−θ)−tan⁡(55∘+θ)+cot⁡(35∘−θ)\cos ec(75^\circ + \theta ) - \sec (15^\circ - \theta ) - \tan (55^\circ + \theta ) + \cot (35^\circ - \theta ) , is
A.-1
B.0
C.1
D.32\dfrac{3}{2}

Explanation

Solution

Hint : The set of two angles whose sum is equal to 90∘90^\circ are called complementary angles. Using trigonometric ratios of complementary angles, we can convert the cosec form into sec form and the cot form into tan form to make the equation simpler and solvable.

Complete step-by-step answer :
The complementary angle of sine is cos –
⇒sin⁡(90∘−θ)=cos⁡θ\Rightarrow \sin (90^\circ - \theta ) = \cos \theta
The complementary angle of cos is sine –
⇒cos⁡(90∘−θ)=sin⁡θ\Rightarrow \cos (90^\circ - \theta ) = \sin \theta
The complementary angle of cosec is sec –
cos⁡ec(90∘−θ)=sec⁡θ\cos ec(90^\circ - \theta ) = \sec \theta
The complementary angle of sec is cosec –
⇒sec⁡(90∘−θ)=cos⁡ecθ\Rightarrow \sec (90^\circ - \theta ) = \cos ec\theta
The complementary angle of tan is cot –
⇒tan⁡(90∘−θ)=cot⁡θ\Rightarrow \tan (90^\circ - \theta ) = \cot \theta
The complementary angle of cot is tan –
cot⁡(90∘−θ)=tan⁡θ\cot (90^\circ - \theta ) = \tan \theta
In this question we have to find
cos⁡ec(75∘+θ)−sec⁡(15∘−θ)−tan⁡(55∘+θ)+cot⁡(35∘−θ)\cos ec(75^\circ + \theta ) - \sec (15^\circ - \theta ) - \tan (55^\circ + \theta ) + \cot (35^\circ - \theta ) .
Now complementary of cos⁡ec(75∘+θ)\cos ec(75^\circ + \theta ) is sec⁡[90∘−(75∘+θ)]=sec⁡(15∘−θ)\sec [90^\circ - (75^\circ + \theta )] = \sec (15^\circ - \theta ) .
So, cos⁡ec(75∘+θ)=sec⁡(15∘−θ)\cos ec(75^\circ + \theta ) = \sec (15^\circ - \theta )
Complementary of cot⁡(35∘−θ)\cot (35^\circ - \theta ) is tan⁡[90∘−(35∘−θ)]=tan⁡(55∘+θ)\tan [90^\circ - (35^\circ - \theta )] = \tan (55^\circ + \theta )
So, cot⁡(35∘−θ)=tan⁡(55∘+θ)\cot (35^\circ - \theta ) = \tan (55^\circ + \theta )
Putting the above two results in the given equation, we get –
sec(15∘−θ)−sec⁡(15∘−θ)−tan⁡(55∘+θ)+tan⁡(55∘+θ)=0sec(15^\circ - \theta ) - \sec (15^\circ - \theta ) - \tan (55^\circ + \theta ) + \tan (55^\circ + \theta ) = 0
So, the correct answer is “0”.

Note : Sine, cosine, tangent, cotangent, cosecant and secant are six main trigonometric ratios,. These six trigonometric ratios are abbreviated as sin, cos, tan, cot, cosec and sec respectively. They are called trigonometric ratios because they are expressed in terms of the ratio of sides of a right angled triangle for a specific angle θ.