Trigonometric Functions

16cos²θ + 25sin²θ + 40sinθ cosθ = 1

16 + 9sin²θ + 20sin 2θ = 1

$16+9\left(\frac{1-cos2\theta }{2}\right)$            + 20sin 2θ = 1

$\frac{-9}{2}cos2\theta +20sin2\theta =\frac{-39}{2}$            

– 9cos 2θ + 40sin 2θ = – 39

$-9\left(\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }\right)+40\left(\frac{2tan\theta }{1+ta{n}^2\theta }\right)=-39$            

48tan²θ + 80tanθ + 30 = 0

24tan²θ + 40tanθ + 15 = 0

  $tan\theta =\frac{-40\pm \sqrt[]{{\left(40\right)}^2-15*24*4}}{2*24}$        

  $tan\theta =\frac{-40\pm \sqrt[]{160}}{2*24}$           

$=\frac{-10\pm \sqrt[]{10}}{12}$            

-> $tan\theta =\frac{\sqrt[]{10}-10}{12}$ , $tan\theta =\frac{-\sqrt[]{10}-10}{12}$  

So $tan\theta =-\frac{\sqrt[]{10}-10}{12}$  will be rejected as $\theta \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$  

Option (4) is correct.

12x =

 $3x=\frac{\pi }{4}$

$cos3x=\frac{1}{\sqrt[]{2}}$

$4co{s}^{3}x-3cosx=\frac{1}{\sqrt[]{2}}$

$4\sqrt[]{2}co{s}^{3}x-3\sqrt[]{2}cosx-1=0$            

$x=\frac{\pi }{12}$ is the solution of above equation.

Statement 1 is true

$f\left(x\right)=4\sqrt[]{2}{x}^3-3\sqrt[]{2}x-1$

$f\text{'}\left(x\right)=12\sqrt[]{2}{x}^2-3\sqrt[]{2}=0$

$x=\pm \frac{1}{2}$

$f\left(-\frac{1}{2}\right)=-\frac{1}{\sqrt[]{2}}+\frac{3}{\sqrt[]{2}}-1=\sqrt[]{2}-1>0$            

f(0) = – 1 < 0

one root lies in $\left(-\frac{1}{2},0\right)$ , one root is $cos\frac{\pi }{12}$  which is positive. As the coefficients are real, therefore all the roots must be real.

Statement 2 is false.

$tanB*tanC=\frac{\sqrt[]{x}}{\sqrt[]{x^2+x+1}}*\frac{1}{\sqrt[]{x\left(x^2+x+1\right)}}$

$=\frac{1}{x^2+x+1}=ta{n}^{2}A$            

tan² A = tan B tan C

It is only possible when A = B = C at x = 1

A = 30°, B = 30°, C = 30° $\left[tanA=tanB=tanC=\frac{1}{\sqrt[]{3}}\right]$                                 

$A+B=\frac{\pi }{2}-C$

a = sin⁻¹ (sin5) = 5 − 2π

and b = cos⁻¹ (cos5) = 2π − 5

∴    a² + b² = (5 − 2π)² + (2π − 5)²

= 8π² − 40π + 50

sin 2 + tan 2 > 0

$\frac{2tan\theta }{1+ta{n}^2\theta }+\frac{2tan\theta }{1-ta{n}^2\theta }>0$

Let tan = x

$\frac{2x}{1+x^2}+\frac{2x}{1-x^2}>0$

$tan\theta <-1or0<tan\theta <1$

$\theta \in \left(0,\frac{\pi }{4}\right)\cup \left(\frac{\pi }{2},\frac{3\pi }{4}\right)\cup \left(\pi ,\frac{5\pi }{4}\right)\cup \left(\frac{3\pi }{2},\frac{7\pi }{4}\right)$

sin x = 1 – sin² x

=> sin x = $\frac{-1+\sqrt[]{5}}{2},\frac{-1-\sqrt[]{5}}{2}\left(rejected\right)$  

draw y = sin x

y = $\frac{\sqrt[]{5}-1}{2},$ find their pt. of intersection.

pva -> $\left(\sim rvp\right)$  

$\sim \left(p\vee q\right)\vee \left(\sim rvp\right)$  

its negation as asked in question

$\left(p\vee q\right)\wedge \left(\sim p\wedge r\right)$  

=  $\left(p\wedge \sim p\wedge r\right)\vee \left(q\wedge r\wedge \sim p\right)$  

=  $\left(p\wedge r\wedge \sim p\right)\left[asp\wedge \sim pisfalse\right]$  

sin x = 1 – sin² x

sin x =  $\frac{-1+\sqrt[]{5}}{2},\frac{-1-\sqrt[]{5}}{2}\left(rejected\right)$  

draw y = sin x

y =  $\frac{\sqrt[]{5}-1}{2},$ find their pt. of intersection.