Question 13 Marks
Prove the following.
(secθ + tanθ) (1 - sinθ) = cosθ
AnswerTaking LHS
$(1-\sin \theta)(\sec \theta+\tan \theta) $
$=(1-\sin \theta)\left(\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}\right) $
$=(1-\sin \theta) \frac{1}{\cos \theta}(1+\sin \theta) $
$=\frac{1}{\cos \theta}\left(1-\sin ^2 \theta\right)\left[(a+b)(a-b)=a^2-b^2\right] $
$=\frac{1}{\cos \theta}\left(\cos ^2 \theta\right)\left[\sin ^2 \theta+\cos ^2 \theta=1\right] $
$=\cos \theta$
$=\text { RHS }$
Proved!
View full question & answer→Question 23 Marks
Prove the following.
secθ(1 - sinθ) (secθ + tanθ) = 1
Answer$\text { Taking LHS }$
$\sec \theta(1-\sin \theta)(\sec \theta+\tan \theta) $
$=\frac{1}{\cos \theta}(1-\sin \theta)\left(\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}\right) $
$=\frac{1}{\cos \theta}(1-\sin \theta) \frac{1}{\cos \theta}(1+\sin \theta) $
$=\frac{1}{\cos ^2 \theta}\left(1-\sin ^2 \theta\right)\left[(a+b)(a-b)=a^2-b^2\right] $
$=\frac{1}{\cos ^2 \theta}\left(\cos ^2 \theta\right)\left[\sin ^2 \theta+\cos ^2 \theta=1\right] $
$=1 $
=\text { RHS }
\text { Proved! }
View full question & answer→Question 33 Marks
A kite is flying at a height of 60 m above the ground. The string attached to the kite is tied at the ground. It makes an angle of 60° with the ground. Assuming that the string is straight, find the length of the string. (√3 = 1.73 )
Answer
Let AC be a string and kite is flying at point A, with height
AB = 60 m
Angle make by string with horizontal, θ = ∠ACB = 60°
Clearly, ABC is a right-angled triangle.
In ∆ABC
$\sin \theta=\frac{A B}{A C}$
$\Rightarrow \sin 60^{\circ}=\frac{60}{A C}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{60}{A C}$
$\Rightarrow A C=\frac{120}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{120 \sqrt{3}}{3}=40 \sqrt{3}$
⇒ AC = 40(1.73)
⇒ AC = 6.92 m
Hence, length of string is 6.92 m View full question & answer→Question 43 Marks
A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60° with the horizontal. Find the height of the tree.
Answer
Let AB be a tree, and C is the point of break.
Height of tree = BC + AC
As, the treetop is rested 20 m from the base, making an angle of 60° with the horizontal.
In ∆ABC
AB = 20 m
∠ABC, θ = 60°
Now,
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}=\frac{ AC }{ AB }$
$\Rightarrow \tan 60^{\circ}=\frac{ AC }{20}$
$\Rightarrow AC =20 \tan 60^{\circ}$
$\Rightarrow AC =20 \sqrt{ } m \text { Base }$
$\cos \theta=\frac{ BB }{\text { Hypotenuse }}=\frac{ AB }{ BC }$
$\Rightarrow \cos 60^{\circ}=\frac{20}{ BC }$
$\Rightarrow \frac{1}{2}=\frac{20}{ BC }$
⇒ BC = 40 m
Height of tree = BC + AC = (40 + 20√3) meters. View full question & answer→Question 53 Marks
Two poles of heights 18 metre and 7 metre are erected on a ground. The length of the wire fastened at their tops in 22 metre. Find the angle made by the wire with the horizontal.
Answer
Let AB and CD be two poles, and AC be the wire joining their top.
and
Angle made by wire with horizontal = ∠CAP = θ [say]
Given,
AB = 7 m [Let AB be smaller pole]
CD = 18 m
AC = 22 m
Clearly, APDB is a rectangle with
AB = PD = 7 m
Also,
CP = CD - PD = 18 - 7 = 11 m
In ∆APC
$\sin \theta=\frac{C P}{A C}$
$\Rightarrow \sin \theta=\frac{11}{22}=\frac{1}{2}$
$\Rightarrow \theta=30^{\circ}\left[\sin 30^{\circ}=\frac{1}{2}\right]$ View full question & answer→Question 63 Marks
Two buildings are facing each other on a road of width 12 metre. From the top of the first building, which is 10 metre high, the angle of elevation of the top of the second is found to be 60°. What is the height of the second building?
Answer
Let AB and CD be two building, with
AB = 10 m
And angle of elevation from top of AB to top of CD = ∠CAP = 60°
Width of road = BD = 12 m
Clearly, ABDP is a rectangle
With
AB = PD = 10 m
BD = AP = 12 m
And APC is a right-angled triangle, In ∆APC
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}=\frac{ CP }{ AP }$
$\Rightarrow \tan 60^{\circ}=\frac{ CP }{12}$
$\Rightarrow \sqrt{3}=\frac{ CP }{12}$
⇒ CP = 12√3 m
Also,
CD = CP + PD = (12√3 + 10) m
Hence, height of other building is (10 + 12√3 m). View full question & answer→Question 73 Marks
From the top of a lighthouse, an observer looking at a ship makes angle of depression of 60°. If the height of the lighthouse is 90 metre, then find how far the ship is from the lighthouse. (√3 = 1.73)
Answer
Let PQ be a light house of height 80 cm such that PQ = 90 m
And R be a ship.
Angle of depression from P to ship R = ∠BPR = 60°
Also, ∠PRQ(say θ) = ∠BPR = 60° [Alternate Angles]
Clearly, PQR is a right-angled triangle.
Now, In ∆PQR
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}=\frac{P Q}{Q R}$
$\Rightarrow \tan 60^{\circ}=\frac{90}{Q R}$
$\Rightarrow \sqrt{3}=\frac{90}{Q R}$
$\Rightarrow Q R=\frac{90}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{90 \sqrt{3}}{3}=30 \sqrt{3}$
⇒ QR = 30(1.73)
⇒ QR = 51.90 m
Hence, Ship is 51.90 m away from the light house. View full question & answer→Question 83 Marks
A person is standing at a distance of 80m from a church looking at its top. The angle of elevation is of 45°. Find the height of the church.
Answer
Let 'A' be the person, standing 80 m away from a church BC,
Angle of elevation, ∠BAC = θ = 45°
Clearly, ∆ABC is a right-angled triangle, in which
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}=\frac{B C}{A B}$
$ \Rightarrow \tan 45^{\circ}=\frac{B C}{80} $
$ \Rightarrow 1=\frac{B C}{80}$
⇒ BC = 80 m
Therefore, Height of church is 80 m. View full question & answer→Question 93 Marks
Prove the following.
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}$
AnswerTaking LHS
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}$
Dividing numerator and denominator by cosθ
$=\frac{\frac{\sin \theta}{\cos \theta}-\frac{\cos \theta}{\cos \theta}+\frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\cos \theta}-\frac{1}{\cos \theta}}$
$=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta} \times \frac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta}$
$=\frac{(\tan \theta-1+\sec \theta)}{(\tan \theta+1-\sec \theta)} \frac{(\sec \theta-\tan \theta)}{(\sec \theta-\tan \theta)}$
$=\frac{\tan \theta \sec \theta-\tan ^2 \theta-\sec \theta+\tan \theta+\sec ^2 \theta-\tan \theta \sec \theta}{(\tan \theta+1-\sec \theta)(\sec \theta-\tan \theta)}$
$=\frac{\tan \theta+1-\sec \theta+\sec ^2 \theta-\tan ^2 \theta}{(\tan \theta+1-\sec \theta)(\sec \theta-\tan \theta)}$
$=\frac{\tan \theta+1-\sec \theta}{(\tan \theta+1-\sec \theta)(\sec \theta-\tan \theta)}\left[\text { As } \sec ^2 \theta-\tan ^2 \theta=1\right]$
$=\frac{1}{\sec \theta-\tan \theta}$
= RHS
Proved.
View full question & answer→Question 103 Marks
Prove the following.
$\frac{\tan ^3 \theta-1}{\tan \theta-1}=\sec ^2 \theta+\tan \theta$
AnswerTaking LHS
$\frac{\tan ^3 \theta-1}{\tan \theta-1}$
$=\frac{(\tan \theta-1)\left(\tan ^2 \theta+\tan \theta+1\right)}{\tan \theta-1}\left[a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=\tan ^2 \theta+\tan \theta+1$
$=\sec ^2 \theta+\tan \theta\left[1+\tan ^2 \theta=\sec ^2 \theta\right]$
$=\text { RHS }$
Proved.
View full question & answer→Question 113 Marks
Prove the following.
$\frac{\tan \theta}{\sec \theta+1}=\frac{\sec \theta-1}{\tan \theta}$
AnswerTaking LHS
$=\frac{\tan \theta}{\sec \theta+1} \times \frac{\sec \theta-1}{\sec \theta-1}$
$=\frac{\tan \theta(\sec \theta-1)}{\sec ^2 \theta-1}$
$=\frac{\tan \theta\left(\sec ^2 \theta-1\right)}{\tan ^2 \theta}\left[\tan ^2 \theta=\sec ^2 \theta-1\right]$
$=\frac{\sec \theta-1}{\tan \theta}$
= RHS
Proved !
View full question & answer→Question 123 Marks
If $\cot \theta=\frac{40}{9},$ find the values of cosecθ and sinθ.
AnswerWe know that,
$\operatorname{cosec}^2 \theta=1+\cot ^2 \theta$
$\Rightarrow \operatorname{cosec}^2 \theta=1+\left(\frac{40}{9}\right)^2$
$\Rightarrow \operatorname{cosec}^2 \theta=1+\frac{1600}{81}$
$\Rightarrow \operatorname{cosec}^2 \theta=\frac{1681}{81}$
$\Rightarrow \operatorname{Cosec} \theta=\frac{41}{9}$
$\Rightarrow$ Also,
$\Rightarrow \sin \theta=\frac{1}{\operatorname{cosec} \theta}$
$\therefore \sin \Theta=\frac{9}{41}$
View full question & answer→Question 133 Marks
Prove the following.
$sec^6x – tan^6x = 1 + 3 sec^2x \times tan^2x$
AnswerTaking LHS
$\sec ^6 x-\tan ^6 x$
$=\left(\sec ^2 x\right)^3-\left(\tan ^2 x\right)^3$
$=\left(\sec ^2 x-\tan ^2 x\right)\left(\sec ^4 x+\tan ^2 x \sec ^2 x+\tan ^4 x\right)$
$\left[A s, a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=\sec ^4 x+\tan ^4 x+\tan ^2 x \sec ^2 x+2 \tan ^2 x \sec ^2 x-2 \tan ^2 x \sec ^2 x$
[As, $\sec ^2 \theta-\tan ^2 \theta=1$ ]
$=\sec ^4 x+\tan ^4 x-2 \tan ^2 x \sec ^2 x+3 \tan ^2 x \sec ^2 x$
$=\left(\sec ^2 x-\tan ^2 x\right)^2+3 \tan ^2 x \sec ^2 x\left[a^2+b^2-2 a b=(a-b)^2\right]$
$=1^2+3 \tan ^2 x \sec ^2 x$
$=1+3 \tan ^2 x \sec ^2 x$
$=\text { RHS }$
Proved.
View full question & answer→Question 143 Marks
If $\tan \theta=\frac{3}{4}$ find the values of secθ and cosθ.
AnswerWe know that,
$\sec ^2 \theta=1+\tan ^2 \theta$
$\Rightarrow \sec ^2 \theta=1+\left(\frac{3}{4}\right)^2$
$\Rightarrow \sec ^2 \theta=1+\frac{9}{16}$
$\Rightarrow \sec ^2 \theta=\frac{25}{16}$
$\Rightarrow \sec \theta=\frac{5}{4}$
Also,
$\Rightarrow \cos \theta=\frac{1}{\sec \theta}$
$\Rightarrow \cos \theta=\frac{4}{5}$
View full question & answer→Question 153 Marks
Prove the following.
$\sec ^2 \theta+\operatorname{cosec}^2 \theta=\sec ^2 \theta \times \operatorname{cosec}^2 \theta$
AnswerTaking LHS
$=\sec ^2 \theta + \operatorname{cosec}^2 \theta$
$=\frac{1}{\cos ^2 \theta}+\frac{1}{\sin ^2 \theta}$
$=\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos ^2 \theta \sin ^2 \theta}$
$ =\frac{1}{\cos ^2 \theta \sin ^2 \theta}\left[\sin ^2 \theta+\cos ^2 \theta=1\right]$
$=\sec ^2 \theta \times \operatorname{cosec}^2 \theta$
$=\text { RHS }$
Proved!
View full question & answer→Question 163 Marks
A tree was broken due to storm. Its broken upper part was so inclined that its top touched the ground making an angle of 30°with the ground.The distance from the foot of the tree and the point where the top touched the ground was 10 metre. What was the height of the tree.
AnswerAs shown in figure 6.13, suppose AB is the tree. It was broken at ‘C’ and its top touched at ‘D’.
$ \angle \mathrm{CDB}=30^{\circ}, \mathrm{BD}=10 \mathrm{~m}, \mathrm{BC}=x \mathrm{~m}$
$\mathrm{CA}=\mathrm{CD}=y \mathrm{~m} $
In right angled $\Delta \mathrm{CDB}$,
$ \tan 30^{\circ}=\frac{\mathrm{BC}}{\mathrm{BD}}$
$\frac{1}{\sqrt{3}}=\frac{x}{10}$
$x=\frac{10}{\sqrt{3}}$
$y=\frac{20}{\sqrt{3}}$
$x+y=\frac{10}{\sqrt{3}}+\frac{20}{\sqrt{3}}$
$=\frac{30}{\sqrt{3}}$
$x+y=10 \sqrt{3}$
$\therefore \text { height of the tree was } 10 \sqrt{3} \mathrm{~m} . $
$\therefore$ height of the tree was $10 \sqrt{3} \mathrm{~m}$.
View full question & answer→Question 173 Marks
From the top of a building, an observer is looking at a scooter parked at some distance away, makes an angle of depression of $30^{\circ}$. If the height of the building is $40 \mathrm{~m}$, find how far the scooter is from the building. $(\sqrt{3}=1.73)$
AnswerIn the figure $6.10, \mathrm{AB}$ is the building. A scooter is at $\mathrm{C}$ which is ' $x$ ' $m$ away from the building.
In figure, ' $A$ ' is the position of the observer.
$\mathrm{AM}$ is the horizontal line and $\angle \mathrm{MAC}$ is the angle of depression. $\angle \mathrm{MAC}$ and $\angle \mathrm{ACB}$ are alternate angles.
from fig,
$
\begin{aligned}
\tan 30^{\circ} & =\frac{A B}{B C} \\
\therefore \quad \frac{1}{\sqrt{3}} & =\frac{40}{x} \\
\therefore \quad x & =40 \sqrt{3} \\
& =40 \times 1.73 \\
& =69.20 \mathrm{~m} .
\end{aligned}
$
$\therefore$ the scooter is $69.20 \mathrm{~m}$. away from the building.
View full question & answer→Question 183 Marks
An observer at a distance of $10 \mathrm{~m}$ from a tree looks at the top of the tree, the angle of elevation is $60^{\circ}$. What is the height of the tree ? $(\sqrt{3}=1.73)$
AnswerIn figure $6.9, \mathrm{AB}=\mathrm{h}=$ height of the tree.
$\mathrm{BC}=10 \mathrm{~m}$, distance of the observer from the tree .
Angle of elevation $(\theta)=\angle \mathrm{BCA}=60^{\circ}$
from figure, $\tan \theta=\frac{\mathrm{AB}}{\mathrm{BC}}$
$\tan 60^{\circ}=\sqrt{3}$
$\therefore \frac{\mathrm{AB}}{\mathrm{BC}}=\sqrt{3}$
from equation (I)and (II)
$\therefore \mathrm{AB}=\mathrm{BC} \sqrt{3}=10 \sqrt{3}$
$\therefore \mathrm{AB}=10 \times 1.73=17.3 \mathrm{~m}$
$\therefore$ height of the tree is $17.3 \mathrm{~m}$.
View full question & answer→Question 193 Marks
Eliminate $\theta$ from given equations.
$ x=a \cot \theta-b \operatorname{cosec} \theta$
$y=a \cot \theta+b \operatorname{cosec} \theta $
Answer$ x=a \cot \theta-b \operatorname{cosec} \theta........1$
$y=a \cot \theta+b \operatorname{cosec} \theta........2 $
Adding equations (I) and (II).
$ x+y=2 a \cot \theta$
$\therefore \cot \theta=\frac{x+y}{2 a} $
Subtracting equation (II) from (I),
$ y-x=2 b \operatorname{cosec} \theta$
$\therefore \operatorname{cosec} \theta=\frac{y-x}{2 b} $
Now, $\operatorname{cosec}^2 \theta-\cot ^2 \theta=1$
$ \therefore\left(\frac{y-x}{2 b}\right)^2-\left(\frac{y+x}{2 a}\right)^2=1$
$\therefore \frac{(y-x)^2}{4 b^2}-\frac{(y+x)^2}{4 a^2}=1$
$\text { or }\left(\frac{y-x}{b}\right)^2-\left(\frac{y+x}{a}\right)^2=4 $
View full question & answer→Question 203 Marks
If $5 \sin \theta-12 \cos \theta=0$, find the values of $\sec \theta$ and $\operatorname{cosec} \theta$..
Answer$ \therefore 5 \sin \theta-12 \cos \theta=0$
$\therefore 5 \sin \theta=12 \cos \theta$
$\therefore \frac{\sin \theta}{\cos \theta}=\frac{12}{5}$
$\therefore \tan \theta=\frac{12}{5} $
we have,
$1+\tan ^2 \theta =\sec ^2 \theta$
$\therefore 1+\left(\frac{12}{5}\right)^2 =\sec ^2 \theta$
$\therefore 1+\frac{144}{25} =\sec ^2 \theta$
$\therefore \quad \frac{25+144}{25} =\sec ^2 \theta$
$\therefore \quad \sec ^2 \theta =\frac{169}{25}$
$\therefore \quad \sec \theta =\frac{13}{5} $
$\therefore \cos \theta=\frac{5}{13}$
Now, $\sin ^2 \theta+\cos ^2 \theta=1$
$\therefore \sin ^2 \theta =1-\cos ^2 \theta$
$\therefore \sin ^2 \theta =1-\left(\frac{5}{13}\right)^2$
$ =1-\frac{25}{169}$
$ =\frac{144}{169}$
$\therefore \sin \theta =\frac{12}{13}$
$\therefore \operatorname{cosec} \theta =\frac{13}{12} $
View full question & answer→Question 213 Marks
For a person standing at a distance of 80 m from a temple, the angle of elevation of its top is 45. Find the height of the church.
Question 223 Marks
Prove : $\frac{\cos ^2 \theta}{1-\tan \theta}+\frac{\sin ^3 \theta}{\sin \theta-\cos \theta}==1+\sin \theta$
View full question & answer→Question 233 Marks
Prove : $\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}$
View full question & answer→Question 243 Marks
Prove : $\frac{\cos ^2 A+\tan ^2 A-1}{\sin ^2 A}=\tan ^2 A$
View full question & answer→Question 253 Marks
Prove : $\left(\frac{1}{\cos \theta}+\frac{1}{\cot \theta}\right)=(\sec \theta-\tan \theta)=1$
View full question & answer→Question 263 Marks
Prove : $\operatorname{cosec}^2 A-\cos ^2 A=\frac{\sec ^2 A-\sin ^2 A}{\tan ^2 A}$
View full question & answer→Question 273 Marks
Prove : $\frac{\sin \theta+\tan \theta}{\cos \theta}=\tan \theta(1+\sec \theta)$
View full question & answer→Question 283 Marks
Prove : $\frac{\tan ^3 A-1}{\tan A -1}=\sec ^2 A+\tan A$
View full question & answer→Question 293 Marks
Prove : $\frac{\cos \theta}{1+\sin \theta}=\sec \theta-\tan \theta$
View full question & answer→Question 303 Marks
Prove : $\sec \theta+\tan \theta=\frac{1}{\sec \theta-\tan \theta}$
View full question & answer→Question 313 Marks
Prove : $\sin ^4 \theta-\cos ^4 \theta=1-2 \cos ^2 \theta$
View full question & answer→Question 323 Marks
Prove : $\cot ^2 \theta-\frac{1}{\sin ^2 \theta}=-1$
View full question & answer→Question 333 Marks
Prove : $\left(1+\cot ^2 \theta\right)(1+\cos \theta)(1-\cos \theta)=1$
View full question & answer→Question 343 Marks
Prove : $\left(1+\tan ^2 \theta\right)(1+\sin \theta)(1-\sin \theta)=1$
View full question & answer→Question 353 Marks
Prove : $\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta}=2 \sec ^2 \theta$
View full question & answer→Question 363 Marks
Prove : $\cos ^2 \theta+\frac{1}{1+\cot ^2 \theta}=1$
View full question & answer→Question 373 Marks
If $\sqrt{3} \tan \theta=3 \sin \theta$, find the value of $\sin ^2 \theta-\cos ^2 \theta$.
Answer$\sin ^2 \theta-\cos ^2 \theta=\frac{1}{2}$
View full question & answer→Question 383 Marks
$3 \sin \theta-4 \cos \theta=0$, then find the values of all trigonometric ratios.
Answer$\sin \theta$ = $\frac{4}{5}$
$\cos \theta$ = $\frac{3}{5}$
$\operatorname { t a n } \theta$ = $\frac{4}{3}$
$\cot \theta$ = $\frac{3}{4}$
$\operatorname { s e c } \theta$ = $\frac{5}{3}$
$\operatorname{cosec} \theta$ = $\frac{5}{4}$
View full question & answer→Question 393 Marks
If cot $\theta$ = $\frac{7}{24}$, find the values of other trigonometric ratios using the identity.
Answer$\sin \theta$ = $\frac{24}{25}$
$\cos \theta$ = $\frac{7}{25}$
$\operatorname { t a n } \theta$ = $\frac{24}{7}$
$\cot \theta$ = $\frac{7}{24}$
$\operatorname { s e c } \theta$ = $\frac{25}{7}$
$\operatorname{cosec} \theta$ = $\frac{25}{25}$
View full question & answer→Question 403 Marks
If tan $\theta$ = 2, find the values of other trigonometric ratios using the identities.
Answer$\sin \theta$ = $\frac{2}{\sqrt{5}}$
$\cos \theta$ = $\frac{1}{\sqrt{5}}$
$\operatorname { t a n } \theta$ = 2
$\cot \theta$ = $\frac{1}{2}$
$\operatorname { s e c } \theta$ = $\sqrt{5}$
$\operatorname{cosec} \theta$ = $\frac{\sqrt{5}}{2}$
View full question & answer→Question 413 Marks
Prove the following.
(secθ + tanθ) (1 - sinθ) = cosθ
Question 423 Marks
Prove the following.
secθ(1 - sinθ) (secθ + tanθ) = 1
Question 433 Marks
A kite is flying at a height of 60 m above the ground. The string attached to the kite is tied at the ground. It makes an angle of 60° with the ground. Assuming that the string is straight, find the length of the string. (√3 = 1.73 )
Question 443 Marks
A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60° with the horizontal. Find the height of the tree.
Question 453 Marks
Two poles of heights 18 metre and 7 metre are erected on a ground. The length of the wire fastened at their tops in 22 metre. Find the angle made by the wire with the horizontal.
Question 463 Marks
Two buildings are facing each other on a road of width 12 metre. From the top of the first building, which is 10 metre high, the angle of elevation of the top of the second is found to be 60°. What is the height of the second building?
Question 473 Marks
From the top of a lighthouse, an observer looking at a ship makes angle of depression of 60°. If the height of the lighthouse is 90 metre, then find how far the ship is from the lighthouse. (√3 = 1.73)
Question 483 Marks
A person is standing at a distance of 80m from a church looking at its top. The angle of elevation is of 45°. Find the height of the church.
Question 493 Marks
Prove the following.
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}$
View full question & answer→Question 503 Marks
Prove the following.
$\frac{\tan ^3 \theta-1}{\tan \theta-1}=\sec ^2 \theta+\tan \theta$
View full question & answer→Question 513 Marks
Prove the following.
$\frac{\tan \theta}{\sec \theta+1}=\frac{\sec \theta-1}{\tan \theta}$
View full question & answer→Question 523 Marks
If $\cot \theta=\frac{40}{9}$, find the values of cosecθ and sinθ.
View full question & answer→Question 533 Marks
Prove the following.
sec6x – tan6x = 1 + 3 sec2x × tan2x
Question 543 Marks
If $\tan \theta=\frac{3}{4}$, find the values of secθ and cosθ.
View full question & answer→Question 553 Marks
Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ
Question 563 Marks
A tree was broken due to storm. Its broken upper part was so inclined that its top touched the ground making an angle of 30°with the ground.The distance from the foot of the tree and the point where the top touched the ground was 10 metre. What was the height of the tree.
Question 573 Marks
From the top of a building, an observer is looking at a scooter parked at some distance away, makes an angle of depression of $30^{\circ}$. If the height of the building is $40 \mathrm{~m}$, find how far the scooter is from the building. $(\sqrt{3}=1.73)$
View full question & answer→Question 583 Marks
An observer at a distance of $10 \mathrm{~m}$ from a tree looks at the top of the tree, the angle of elevation is $60^{\circ}$. What is the height of the tree ? $(\sqrt{3}=1.73)$
View full question & answer→Question 593 Marks
Eliminate $\theta$ from given equations.
x=a $\cot$ $\theta$ -b cosec $\theta$
y=a $\cot$ $\theta$+b cosec $\theta$
View full question & answer→Question 603 Marks
If $5 \sin \theta-12 \cos \theta=0$, find the values of $\sec \theta$ and $\operatorname{cosec} \theta$..
View full question & answer→Question 613 Marks
Prove : $\sin ^4 \theta-\cos ^4 \theta=1-2 \cos ^2 \theta$
View full question & answer→Question 623 Marks
Prove : $\sec \theta+\tan \theta=\frac{1}{\sec \theta-\tan \theta}$
View full question & answer→Question 633 Marks
Prove : $\operatorname{cosec}^2 A-\cos ^2 A=\frac{\sec ^2 A-\sin ^2 A}{\tan ^2 A}$
View full question & answer→Question 643 Marks
Prove : $\left(\frac{1}{\cos \theta}+\frac{1}{\cot \theta}\right)=(\sec \theta-\tan \theta)=1$
View full question & answer→Question 653 Marks
Prove : $\left(1+\tan ^2 \theta\right)(1+\sin \theta)(1-\sin \theta)=1$
View full question & answer→Question 663 Marks
Prove : $\left(1+\cot ^2 \theta\right)(1+\cos \theta)(1-\cos \theta)=1$
View full question & answer→Question 673 Marks
Prove : $\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}$
View full question & answer→Question 683 Marks
Prove : $\frac{\tan ^3 A-1}{\tan A -1}=\sec ^2 A+\tan A$
View full question & answer→Question 693 Marks
Prove : $\frac{\sin \theta+\tan \theta}{\cos \theta}=\tan \theta(1+\sec \theta)$
View full question & answer→Question 703 Marks
Prove : $\frac{\cos \theta}{1+\sin \theta}=\sec \theta-\tan \theta$
View full question & answer→Question 713 Marks
Prove : $\frac{\cos ^2 \theta}{1-\tan \theta}+\frac{\sin ^3 \theta}{\sin \theta-\cos \theta}==1+\sin \theta$
View full question & answer→Question 723 Marks
Prove : $\frac{\cos ^2 A+\tan ^2 A-1}{\sin ^2 A}=\tan ^2 A$
View full question & answer→Question 733 Marks
Prove : $\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta}=2 \sec ^2 \theta$
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Prove : $\cot ^2 \theta-\frac{1}{\sin ^2 \theta}=-1$
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Prove : $\cos ^2 \theta+\frac{1}{1+\cot ^2 \theta}=1$
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If tan $\theta$ = 2, find the values of other trigonometric ratios using the identities.
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If $\sqrt{3} \tan \theta=3 \sin \theta$, find the value of $\sin ^2 \theta-\cos ^2 \theta$.
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If cot $\theta$ = $\frac{7}{24}$, find the values of other trigonometric ratios using the identity.
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For a person standing at a distance of 80 m from a temple, the angle of elevation of its top is 45. Find the height of the church.
Question 803 Marks
$3 \sin \theta-4 \cos \theta=0$, then find the values of all trigonometric ratios.
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