Question

(a+2) sinα +(2a-1) cosα = (2a+1) if tanα =

(a) 3/4
(b) 4/3
(c) (2a)/(a²+1)
(d) (2a)/(a²-1)

Answer 1

AnsWer :

(b) and (d)

Solution :

We have,

$\tt \dagger \: \: ( \alpha + 2) \sin \alpha + (2 \alpha - 1) \cos \alpha = (2 \alpha + 1)$

Let's tan alpha be X.

Now,

Dividing Cos alpha,We get.

$\tt\mapsto\frac{( \alpha + 2) \sin\alpha}{\cos} + \frac{ (2 \alpha - 1)\cos\alpha }{\cos\alpha }= \frac{(2 \alpha + 1)}{\cos\alpha }.$

Now, Taking alpha term and x term one side and Simplify.

$\rule{90}1$

$\tt \mapsto(3x - 4)(1 - { \alpha }^{2} )( x + 2 \alpha ) = 0.$

$\tt\mapsto \: x = \frac{4}{3} \: \: \: \: \: or \: \: \: \frac{ - 2 \alpha }{1 - {\alpha }^{2} }$

$\tt \mapsto x = \frac{4}{3} \: \: \: \: or \: \: \: \: \: \frac{2 \alpha }{ { \alpha }^{2} - 1 }$

$\rule{200}1$

$\boxed{\begin{minipage}{6 cm}Fundamental Trigonometric Identities \\ \\ \sin^2\theta + \cos^2\theta=1 \\ \\1+\tan^2\theta = \sec^2\theta \\ \\1 +\cot^2\theta = \text{cosec}^2 \, \theta \end{minipage}}$

Note : Whole solution provide above.

Answer 2

Given:

(a+2)sinα+(2a-1)cosα= 2a+1

To Find:

Value of tanα

Solution:

We know that,

$\leadsto\bf{tan\theta=\dfrac{sin\theta}{cos\theta}}$

Now,

It is given that,

$\longrightarrow\mathrm{(a+2)sin\alpha+(2a-1)cos\alpha=2a+1}$

$\longrightarrow\mathrm{asin\alpha+2sin\alpha+2acos\alpha-cos\alpha=2a+1}$

$\longrightarrow\mathrm{asin\alpha+2acos\alpha+2sin\alpha-cos\alpha=2a+1}$

$\longrightarrow\mathrm{a(sin\alpha+2cos\alpha)+2sin\alpha-cos\alpha=2a+1}$

On comparing coefficient of 'a' and constant term of both the sides, we get

$\longrightarrow\mathrm{sin\alpha+2cos\alpha=2}------(1)$

$\longrightarrow\mathrm{2sin\alpha-cos\alpha=1}------(2)$

On multiplying eqⁿ (2) by 2, we get

$\longrightarrow\mathrm{4sin\alpha-2cos\alpha=2}------(3)$

On adding (1) and (3), we get

$\longrightarrow\mathrm{sin\alpha+2cos\alpha+4sin\alpha-2cos\alpha=2+2}$

$\longrightarrow\mathrm{sin\alpha+4sin\alpha+2cos\alpha-2cos\alpha=4}$

$\longrightarrow\mathrm{5sin\alpha=4}$

$\longrightarrow\mathrm{sin\alpha=\dfrac{4}{5}}$

On putting value of sinα in (1), we get

$\longrightarrow\mathrm{\dfrac{4}{5}+2cos\alpha=2}$

$\longrightarrow\mathrm{2cos\alpha=2-\dfrac{4}{5}}$

$\longrightarrow\mathrm{2cos\alpha=\dfrac{10-4}{5}}$

$\longrightarrow\mathrm{\cancel{2}cos\alpha=\dfrac{\cancel{6}}{5}}$

$\longrightarrow\mathrm{cos\alpha=\dfrac{3}{5}}$

Now,

$\longrightarrow\mathrm{tan\alpha=\dfrac{sin\alpha}{cos\alpha}}$

$\longrightarrow\mathrm{tan\alpha=\dfrac{\dfrac{4}{\cancel{5}}}{\dfrac{3}{\cancel{5}}}}$

$\longrightarrow\mathrm{\pink{tan\alpha=\dfrac{4}{3}}}$

Hence, the correct answer is option (b).