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Question: If \[8\operatorname{tanA}=15\], then find \[\sin A-\operatorname{cosA}\]....

If 8tanA⁡=158\operatorname{tanA}=15, then find sin⁡A−cosA⁡\sin A-\operatorname{cosA}.

Explanation

Solution

Hint:First of all we will find the value of sec⁡A\sec A by using 1+tan⁡2A=sec⁡2A1+{{\tan }^{2}}A={{\sec }^{2}}A. Now find cos⁡A\cos A by using cos⁡A=1sec⁡A\cos A=\dfrac{1}{\sec A}. Also find sin⁡A\sin A by using sin2A+cos⁡2A=1si{{n}^{2}}A+{{\cos }^{2}}A=1. Now substitute these values in the given expression to get the required answer.

Complete step-by-step answer:
Here, we are given that 8tanA⁡=158\operatorname{tanA}=15 and we have to find the value of sin⁡A−cosA⁡\sin A-\operatorname{cosA}.
Let us consider the expression in the question.
E=sin⁡A−cosA⁡......(1)E=\sin A-\operatorname{cosA}......(1)
We are given that 8tanA⁡=158\operatorname{tanA}=15.
So we get tanA⁡=158\operatorname{tanA}=\dfrac{15}{8}.
We know that 1+tan⁡2A=sec⁡2A1+{{\tan }^{2}}A={{\sec }^{2}}A.
By substituting the value of tan⁡A\tan A, we get as follows:

& 1+{{\left( \dfrac{15}{8} \right)}^{2}}={{\sec }^{2}}A \\\ & 1+\dfrac{225}{64}={{\sec }^{2}}A \\\ & \dfrac{64+225}{64}={{\sec }^{2}}A \\\ & \dfrac{289}{64}={{\sec }^{2}}A \\\ \end{aligned}$$ So we get $$\sec A=\sqrt{\dfrac{289}{64}}=\dfrac{17}{8}$$. We know that $$\cos A=\dfrac{1}{\sec A}$$. So we get $$\cos A=\dfrac{1}{\dfrac{17}{8}}=\dfrac{8}{17}......(2)$$ We also know that $$si{{n}^{2}}A+{{\cos }^{2}}A=1$$. By substituting the value of $$\operatorname{cosA}=\dfrac{8}{17}$$, we get as follows: $$\begin{aligned} & si{{n}^{2}}A+{{\left( \dfrac{8}{17} \right)}^{2}}=1 \\\ & si{{n}^{2}}A=1-\dfrac{64}{289} \\\ & si{{n}^{2}}A=\dfrac{289-64}{289} \\\ & sinA=\sqrt{\dfrac{225}{289}} \\\ & \sin A=\dfrac{15}{17}......(3) \\\ \end{aligned}$$ Now by substituting $$\sin A=\dfrac{15}{17}$$ and $$\operatorname{cosA}=\dfrac{8}{17}$$ from equation (3) and equation (2) in equation (1) we get as follows: $$\begin{aligned} & E=\sin A-\operatorname{cosA} \\\ & E=\dfrac{15}{17}-\dfrac{8}{17} \\\ & E=\dfrac{15-8}{17} \\\ & E=\dfrac{7}{17} \\\ \end{aligned}$$ So we have got the value of $$\sin A-\operatorname{cosA}$$ as $$\dfrac{7}{17}$$. Note: Students can also solve this question by considering a triangle ABC as follows: ![](https://www.vedantu.com/question-sets/6361a287-671f-41d7-8c0e-42bc59e9d4263244198497969368315.png) We know that $$\operatorname{tanA}=\dfrac{P}{B}=\dfrac{BC}{AB}=\dfrac{15}{8}$$. So take BC as 15x and AB as 3x and use Pythagoras theorem to find AC as 17x. Now find $$\sin A$$ and $$\cos A$$ by using $$\dfrac{P}{H}$$ and $$\dfrac{B}{H}$$ respectively.Substitute the values in the given expression to find the answer.