Introduction to Trigonometry Test - 17

(sin30° + cos30°) – (sin 60° + cos60°)

= $\frac{1}{2}+\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}-\frac{1}{2}$

= 0

Therefore, option A is correct.

Given: $\tan \theta +\cot \theta =2$

$\Rightarrow \tan \theta +\frac{1}{\tan \theta }=2$

$\Rightarrow {\tan }^2\theta +1=2\tan \theta$

$\Rightarrow {\tan }^2\theta -2\tan \theta +1=0$

$\Rightarrow (\tan \theta -1{)}^2=0$

$\Rightarrow \tan \theta -1=0$

$\Rightarrow \tan \theta =1$

Now, $\cot \theta =\frac{1}{\tan \theta }=\frac{1}{1}=1$

$\begin{array}{l}\therefore {\tan }^{25}\theta +{\cot }^{25}\theta =(\tan \theta {)}^{25}+(\cot \theta {)}^{25}=(1{)}^{25}+(1{)}^{25}\end{array}$

$\Rightarrow 1+1=2$

Hence, the correct option is (C).

Given, $\sin A=\frac{a}{b}$

As $\sin A = \frac{Perpendicular }{Hypotenuse}$

Let the complete ratio be $x$

$\Rightarrow Perpendicular = ax$

$Hypotenuse = bx$

Now

${\left(Hypotenuse\right)}^2={\left(Base\right)}^2+{\left(Perpendicular\right)}^2$

${\left(Base\right)}^2={\left(Hypotenuse\right)}^2-{\left(Perpendicular\right)}^2$

${\left(Base\right)}^2={\left(bx\right)}^2-{\left(ax\right)}^2$

$Base= x\sqrt{b^2-a^2}$

$\cos A=\frac{Base}{Hypotenuse}=\frac{x\sqrt{b^2-a^2}}{bx}=\frac{\sqrt{b^2-a^2}}{b}$

Therefore, option C is correct.

$\frac{1}{sec \theta }$ (Given)

$sec \theta$ is in the denominator

The minimum value of $sec \theta$ will return maximum value for $\frac{1}{sec \theta }$

But the minimum value of $sec \theta$ is $sec 0°=1$

Hence, the maximum value of $\frac{1}{sec 0°}=\frac{1}{1}=1$

Hence, the correct option is (C).

Let $AB = 5k$ and $AC = 13k$

If $\sin \frac{5}{13}$

Then the ration of $\frac{\text{ opposite side }}{\text{ hypotenuse }}=\frac{5}{13}$

By Pythagorean Theorem

If opposite side $=5$ units and hypotenuse $=13$ units (for any units)

Then adjacent side $=12$ units

and $\cos =\frac{\text{ adjacent side }}{\text{ hypotenuse }}=\frac{12}{13}$

Hence, the correct option is (C).

Let $AB$ be lighthouse $= 20\sqrt{3}$

Let $C$ be the initial position of the boat

In triangle $ABC$

$\tan {30}^o=\frac{AB}{BC}=\frac{20\sqrt{3}}{BC}$

$\frac{1}{\sqrt{3}}=\frac{20\sqrt{3}}{BC}$

$\Rightarrow BC = 60 m$

In triangle $ABD$

$\tan {60}^o=\frac{20\sqrt{3}}{BD}$

$BD = 20 m$

$CD = 40 m$

Speed of boat = $\frac{CD}{10}=\frac{40}{10}=4m/sec$

Time is taken by boat to reach the lighthouse = $\frac{60}{4}$

$= 15$ sec

Therefore, option C is correct.

cosec (75° + θ) – sec (15° - θ) – tan (55° + θ) + cot (35° - θ)

= cosec (75° + θ) – cosec [90° - (15° - θ)] – tan (55° + θ) + tan [90° - (35° - θ)]

= cosec (75° + θ) – cosec (75° + θ) – tan (55° + θ) + tan (55° + θ)

= 0

Therefore, option  C is correct.

cos θ = 4/5

Base=4; Hypotenuse = 5; Perpendicular = 3

Now, cosecθ = H/P = 5/3; cotθ = B/P = 4/3

So, $\frac{\cos ec\theta }{1+cot\theta }=\left(\frac{{\displaystyle \frac{5}{3}}}{1+{\displaystyle \frac{4}{3}}}\right)$

= $\frac{5}{7}$

Therefore, otpion A is correct.

Given: tanθ = $\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }$

By componendo And dividendo:

∴ $\frac{1+\tan \theta }{1-\tan \theta }=\frac{2\sin \alpha }{2\cos \alpha }=\frac{\sin \alpha }{\cos \alpha }$

⇒ sin α = k (1 + tanθ) and cos α = k(1 – tanθ)

∴ sin2α α + cos2 α = 2k2(1 + tan2θ) = 1

∴ k = ± $\frac{\cos \theta }{\sqrt{2}}$

∴ sin α + cos α = 2k = ±$\sqrt{2\cos \theta }$

Therefore, option A is correct.

Let QR = x cm, then PR = (25−x) cm

$\therefore (25-x{)}^2=x^2+5^2$

$\Rightarrow 625-50x+x^2=x^2+25$

$\Rightarrow -50x=-600$

$\Rightarrow x=12cm$

$\therefore QR=12cm$ and $PR=25-12=13 cm$

Now, $\sin P+\tan P$

$=\frac{12}{13}+\frac{12}{5}=\frac{60+156}{65}=\frac{216}{65}$

hence, the correct option is (A).

If $A, B$ and $C$ are interior angles of a triangle then sum of angles is $180°$

$A+B+C=\pi$

$\Rightarrow B+C=\pi -A$

$\tan \left(\frac{B+C}{2}\right)=tan\left(\frac{\pi -A}{2}\right)=tan\left(\frac{\pi }{2}-\frac{A}{2}\right)$

We know that $\tan (90-\theta )=\cot \theta$

$\therefore \tan \left(\frac{\pi }{2}-\frac{A}{2}\right)=\cot \left(\frac{A}{2}\right)$

Hence, the correct option is (B).

$2 {\sin }^2\theta +3 {\cos }^2\theta$

$=2{\sin }^2\theta +2{\cos }^2\theta +{\cos }^2\theta$

$=2\left({\sin }^2\theta +{\cos }^2\theta \right)+{\cos }^2\theta$  [ identity ${\sin }^2\theta +{\cos }^2\theta =1$ ]

$= 2 +{\cos }^2\theta$

Since we know that $-1\le \cos \theta \le 1$ then $0\le {\cos }^2\theta \le 1$

∵ minimum value of ${\cos }^2\theta = 0$

∴ Required minimum value $= 2 + 0 =2$

Therefore, option B is correct.

tanθ = $\frac{1}{\sqrt{11}}$

Then, cosecθ = $\sqrt{12}$

and secθ = $\frac{\sqrt{12}}{\sqrt{11}}$

Now, $\frac{\cos e{c}^2\theta -se{c}^2\theta }{\cos e{c}^2\theta +se{c}^2\theta }$

= $\frac{\left(12-{\displaystyle \frac{12}{11}}\right)}{\left(12+{\displaystyle \frac{12}{11}}\right)}$

= $\frac{120}{144}=\frac{5}{6}$

Therefore, option B is correct.

Given: $\cos \frac{A}{2}=x$

We know: $\cos A=1-2 {\sin }^2 \frac{A}{2}$

$\Rightarrow \cos A=1-2(1-{\cos }^2\frac{A}{2})$

$\Rightarrow \cos A=1-2+2{\cos }^2\frac{A}{2}$

$\Rightarrow \cos A=-1+2x^2$

$=2x^2=1+\cos A$

$\Rightarrow x^2=\frac{1+\cos A}{2}$

$\Rightarrow x=\sqrt{\frac{1+\cos A}{2}}$

Hence, the correct option is (D).

sec θ + tan θ = 2            ………….. (i)

sec²θ – tan²θ = 1

⟹ (sec θ + tanθ)(sec θ - tanθ) = 1

⟹ sec θ – tan θ = $\frac{1}{2}$        …………(ii)

By adding equations (i) and (ii),

Sec θ + tan θ + sec θ – tan θ = 2 + $\frac{1}{2}$ = $\frac{5}{2}$

⇒ 2secθ = $\frac{5}{2}$

⇒ secθ = $\frac{5}{4}$

Therefore, option B is correct.