If \[7\sin \alpha = 24\cos \alpha \] ; \[0 < \alpha \dfrac{\pi }{2}\] , then value of \[14\tan \alpha - 75\cos \alpha - 7\sec \alpha \] is equal to
A. 1
B. 2
C. 3
D. 4
Answer
612k+ views
Hint: Here the question is related to the trigonometry where it contains the trigonometry ratios like sine, cosecant, cosine, secant tangent and cotangent function. Hence by applying the definition of the trigonometry ratios we can find the result for the given question.
Complete step by step solution:
In trigonometry we have 6 trigonometry ratios namely, sine cosine, tangent, cosecant, secant and cotangent. The ratios are interlinked to each other. The cosecant trigonometry ratio is reciprocal of the sine. The secant trigonometry ratio is reciprocal of the cosine. The cotangent trigonometry ratio is reciprocal of the tangent.
The question is given as \[7\sin \alpha = 24\cos \alpha \] ---- (1)
Divide the above equation by \[\cos \alpha \]
\[7\tan \alpha = 24\] ---- (2)
Divide the equation (2) by 7 we have
\[\tan \alpha = \dfrac{{24}}{7}\]
By the considering the triangle we can define the \[\tan x\]
Let us consider the right-angled triangle ABC
\[\tan x = \dfrac{{opposite}}{{adjacent}} = \dfrac{{AC}}{{BC}}\]
\[ \Rightarrow \dfrac{{24}}{7} = \dfrac{{AC}}{{BC}}\]
Therefore the \[\tan \alpha = \dfrac{{24}}{7}\] ---- (3)
Therefore the value of AC=24 and BC=7
By applying the Pythagoras theorem we have
\[
A{B^2} = A{C^2} + B{C^2} \\
\Rightarrow A{B^2} = {(24)^2} + {(7)^2} \\
\Rightarrow A{B^2} = 576 + 49 \\
\Rightarrow A{B^2} = 625 \\
\Rightarrow AB = \sqrt {625} \;
\]
Hence the length of the AB= \[25\]
By the considering the triangle we can define the \[\cos x\]
\[\cos x = \dfrac{{adjacent}}{{opposite}} = \dfrac{{BC}}{{AB}}\]
\[ \Rightarrow \dfrac{7}{{25}} = \dfrac{{BC}}{{AB}}\]
Therefore the \[\cos \alpha = \dfrac{7}{{25}}\] ----- (4)
The secant trigonometry ratio is reciprocal of cosine. So we have
\[\sec \alpha = \dfrac{1}{{\cos \alpha }} = \dfrac{{25}}{7}\]
Therefore the \[\sec \alpha = \dfrac{{25}}{7}\] ----- (5)
So here we have to find the value of
\[14\tan \alpha - 75\cos \alpha - 7\sec \alpha \]
Substituting the equation (3) to equation (5) to the above inequality we have
\[ \Rightarrow 14\left( {\dfrac{{24}}{7}} \right) - 75\left( {\dfrac{7}{{25}}} \right) - 7\left( {\dfrac{{25}}{7}} \right)\]
On simplifying we have
\[
\Rightarrow 2 \times 24 - 3 \times 7 - 25 \\
\Rightarrow 48 - 21 - 25 \\
\Rightarrow 2 \;
\]
Therefore, the option B is the correct one.
So, the correct answer is “Option B”.
Note: The sine function, cosine function and tan function are defined by considering the right-angled triangle. Hence, we know about the Pythagoras theorem and to which type of triangle it is applicable. For the further simplification we use simple arithmetic operations and hence we get desired results.
Complete step by step solution:
In trigonometry we have 6 trigonometry ratios namely, sine cosine, tangent, cosecant, secant and cotangent. The ratios are interlinked to each other. The cosecant trigonometry ratio is reciprocal of the sine. The secant trigonometry ratio is reciprocal of the cosine. The cotangent trigonometry ratio is reciprocal of the tangent.
The question is given as \[7\sin \alpha = 24\cos \alpha \] ---- (1)
Divide the above equation by \[\cos \alpha \]
\[7\tan \alpha = 24\] ---- (2)
Divide the equation (2) by 7 we have
\[\tan \alpha = \dfrac{{24}}{7}\]
By the considering the triangle we can define the \[\tan x\]
Let us consider the right-angled triangle ABC
\[\tan x = \dfrac{{opposite}}{{adjacent}} = \dfrac{{AC}}{{BC}}\]
\[ \Rightarrow \dfrac{{24}}{7} = \dfrac{{AC}}{{BC}}\]
Therefore the \[\tan \alpha = \dfrac{{24}}{7}\] ---- (3)
Therefore the value of AC=24 and BC=7
By applying the Pythagoras theorem we have
\[
A{B^2} = A{C^2} + B{C^2} \\
\Rightarrow A{B^2} = {(24)^2} + {(7)^2} \\
\Rightarrow A{B^2} = 576 + 49 \\
\Rightarrow A{B^2} = 625 \\
\Rightarrow AB = \sqrt {625} \;
\]
Hence the length of the AB= \[25\]
By the considering the triangle we can define the \[\cos x\]
\[\cos x = \dfrac{{adjacent}}{{opposite}} = \dfrac{{BC}}{{AB}}\]
\[ \Rightarrow \dfrac{7}{{25}} = \dfrac{{BC}}{{AB}}\]
Therefore the \[\cos \alpha = \dfrac{7}{{25}}\] ----- (4)
The secant trigonometry ratio is reciprocal of cosine. So we have
\[\sec \alpha = \dfrac{1}{{\cos \alpha }} = \dfrac{{25}}{7}\]
Therefore the \[\sec \alpha = \dfrac{{25}}{7}\] ----- (5)
So here we have to find the value of
\[14\tan \alpha - 75\cos \alpha - 7\sec \alpha \]
Substituting the equation (3) to equation (5) to the above inequality we have
\[ \Rightarrow 14\left( {\dfrac{{24}}{7}} \right) - 75\left( {\dfrac{7}{{25}}} \right) - 7\left( {\dfrac{{25}}{7}} \right)\]
On simplifying we have
\[
\Rightarrow 2 \times 24 - 3 \times 7 - 25 \\
\Rightarrow 48 - 21 - 25 \\
\Rightarrow 2 \;
\]
Therefore, the option B is the correct one.
So, the correct answer is “Option B”.
Note: The sine function, cosine function and tan function are defined by considering the right-angled triangle. Hence, we know about the Pythagoras theorem and to which type of triangle it is applicable. For the further simplification we use simple arithmetic operations and hence we get desired results.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

