Solveeit Logo
Login

Question

Question: If \[\sec A=\dfrac{17}{8}\], verify that \[\dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3-...

If sec⁡A=178\sec A=\dfrac{17}{8}, verify that
3−4sin⁡2A4cos⁡2A−3=3−tan⁡2A1−3tan⁡2A\dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}

Explanation

Solution

Hint:First of all, consider a triangle ABC, right-angled at B. Now use sec⁡A=HypotenuseBase=178\sec A=\dfrac{Hypotenuse}{Base}=\dfrac{17}{8} and let hypotenuse and base be 17x and 8x respectively. Now, find the remaining side by using the Pythagoras theorem. Now find, sin⁡A=PH,cos⁡A=BH\sin A=\dfrac{P}{H},\cos A=\dfrac{B}{H} and tan⁡A=PB\tan A=\dfrac{P}{B} and substitute in the given equation to prove that LHS = RHS.

Complete step-by-step answer:
Here, we are given that sec⁡A=178\sec A=\dfrac{17}{8}. We have to verify that 3−4sin⁡2A4cos⁡2A−3=3−tan⁡2A1−3tan⁡2A\dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}.
We are given that,
sec⁡A=178....(i)\sec A=\dfrac{17}{8}....\left( i \right)
We know that,
sec⁡A=HypotenuseBase....(ii)\sec A=\dfrac{Hypotenuse}{Base}....\left( ii \right)
From equation (i) and (ii), we get,
178=HypotenuseBase\dfrac{17}{8}=\dfrac{Hypotenuse}{Base}
Let us consider a triangle ABC, right-angled at B.

Let base AB be equal to 8x and hypotenuse be equal to 17x.
We know that Pythagoras theorem states that in a right-angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the other two sides. So, in the above triangle ABC, by applying Pythagoras theorem, we get,
(AB)2+(BC)2=(AC)2{{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}={{\left( AC \right)}^{2}}
Now by substituting the value of AB = 8x and AC = 17x, we get,
(8x)2+(BC)2=(17x)2{{\left( 8x \right)}^{2}}+{{\left( BC \right)}^{2}}={{\left( 17x \right)}^{2}}
64x2+(BC)2=289x264{{x}^{2}}+{{\left( BC \right)}^{2}}=289{{x}^{2}}
BC2=289x2−64x2B{{C}^{2}}=289{{x}^{2}}-64{{x}^{2}}
BC2=225x2B{{C}^{2}}=225{{x}^{2}}
BC=15xBC=15x
Now, we know that,
sin⁡A=PerpendicularHypotenuse\sin A=\dfrac{Perpendicular}{Hypotenuse}
cos⁡A=BaseHypotenuse\cos A=\dfrac{Base}{Hypotenuse}
tan⁡A=PerpendicularBase\tan A=\dfrac{Perpendicular}{Base}
We can see that with respect to angle A,
Perpendicular = BC = 15x
Base = AB = 8x
Hypotenuse = AC = 17x
So, we get,
sin⁡A=BCAC=15x17x=1517\sin A=\dfrac{BC}{AC}=\dfrac{15x}{17x}=\dfrac{15}{17}
cos⁡A=ABAC=8x17x=817\cos A=\dfrac{AB}{AC}=\dfrac{8x}{17x}=\dfrac{8}{17}
tan⁡A=BCAB=15x8x=158\tan A=\dfrac{BC}{AB}=\dfrac{15x}{8x}=\dfrac{15}{8}
Now, let us consider the equation given in the question,
3−4sin⁡2A4cos⁡2A−3=3−tan⁡2A1−3tan⁡2A\dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}
By substituting the value of sin A, cos A and tan A, we get,
3−4(1517)24(817)2−3=3−(158)21−3(158)2\dfrac{3-4{{\left( \dfrac{15}{17} \right)}^{2}}}{4{{\left( \dfrac{8}{17} \right)}^{2}}-3}=\dfrac{3-{{\left( \dfrac{15}{8} \right)}^{2}}}{1-3{{\left( \dfrac{15}{8} \right)}^{2}}}
3−4(225289)4(64289)−3=3−225641−3(22564)\dfrac{3-4\left( \dfrac{225}{289} \right)}{4\left( \dfrac{64}{289} \right)-3}=\dfrac{3-\dfrac{225}{64}}{1-3\left( \dfrac{225}{64} \right)}
3(289)−4(225)4(64)−3(289)=(64)(3)−(225)64−3(225)\dfrac{3\left( 289 \right)-4\left( 225 \right)}{4\left( 64 \right)-3\left( 289 \right)}=\dfrac{\left( 64 \right)\left( 3 \right)-\left( 225 \right)}{64-3\left( 225 \right)}
367−900256−867=192−22564−675\dfrac{367-900}{256-867}=\dfrac{192-225}{64-675}
−33−611=−33−611\dfrac{-33}{-611}=\dfrac{-33}{-611}
33611=33611\dfrac{33}{611}=\dfrac{33}{611}
Hence, LHS = RHS
Therefore, we have verified that
3−4sin⁡2A4cos⁡2A−3=3−tan⁡2A1−3tan⁡2A\dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}

Note: In these types of questions, students must remember to take all trigonometric ratios positive because when nothing is specified about the angle, we consider it to be acute or in the first quadrant. Also, students can find cos A by using cos⁡A=1sec⁡A\cos A=\dfrac{1}{\sec A}, tan A by using tan⁡2A=sec⁡2A−1{{\tan }^{2}}A={{\sec }^{2}}A-1 and sin A by using sin⁡Acos⁡A=tan⁡A\dfrac{\sin A}{\cos A}=\tan A and then substitute these values in the given equation to verify it.