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10th Standard
Maths
If sin (A - B) = \(\frac12\), cos (A + B) = \(\frac12\), 0o < A + ≤ 90°, A > B, find A and B.
Express the ratios cos A, tan A and sec A in terms of sin A.
Evaluate \(\frac { tan{ 65 }^{ o } }{ tan{ 25 }^{ o } } \)
If sin 3A = cos (A - 26°), where 3A is an acute angle, find the value at A.
Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
From a point on a bridge across a river, the angles of depression of the banks on opposite sides at the river are 30° and 45°, respectively. If the bridge is at a height at 3 m from the banks, find the width at the river.
If ATB=90o then prove that
\(\sqrt { \frac { tanA\quad tanB+tanA\quad cotB }{ sinA\quad secB } } -\frac { { Sin }^{ 2 }A }{ { Cos }^{ 2 }A } =tanA\)
P.T (1+tan∝tan∝tanβ)2 +(tan∝-tanβ)2 =sec2 ∝sec2β.
If 15tan2 θ+4 sec2 θ=23 then find the value of (secθ+cosecθ)2 -sin2 θ
\(P.T\left( \frac { 1+{ tan }^{ 2 }A }{ 1+{ cot }^{ 2 }A } \right) ={ \left( \frac { 1-tan\quad A }{ 1-cot\quad A } \right) }^{ 2 }={ tan }^{ 2 }A\)
If tanθ+sinθ=P; tanθ-sinθ=q P.T P2-q2=4\(\sqrt{pq}\)
The angle of elevation of a tower at a point is 45o, After going 20 meters towards the foot of the tower the angle of elevation of the tower becomes 60o calculate the height of the tower.
The shadow of a tower, when the angle of elevation of the sum is 45o is found to be 10 metres, longer than when it is 60o. find the height of the tower
Prove the trigonometric identity
\(\frac{\sin \theta}{1-\cos \theta}=\operatorname{cosec} \theta+\cot \theta\)
Prove the trigonometric identity
\(\frac{\tan \theta+\sin \theta}{\tan \theta-\sin \theta}=\frac{\sec \theta+1}{\sec \theta-1}\)
Answers
Since, sin(A - B) = \(\frac12\), ∴ A-B = 30° ..... (1)
Also, since cos (A + B) = \(\frac12\)
∴ A + B = 60° ...(2)
Solving (1) and (2)
A - B + A + B = 30o + 60o
2A = 90o
A = 45o
We get,
A = 45° and B = 15°
Since
cos2A + sin2A = 1 therefore
cos2A = 1 - sin2A
i.e., cos A = 土 \(\sqrt { 1-{ sin }^{ 2 }A } \)
This gives cos A = \(\sqrt { 1-{ sin }^{ 2 }A } \)
Hence, \(tanA=\frac { sinA }{ cosA } =\frac { sinA }{ \sqrt { { 1-sin }^{ 2 }A } } \)
and \(secA=\frac { 1 }{ cosA } =\frac { 1 }{ \sqrt { 1-{ sin }^{ 2 }A } } \)
We know:
cot A = tan(90o - A)
So,
cot 25° = tan (90° - 25°) = tan 65°
\(\frac { tan{ 65 }^{ o } }{ tan{ 25 }^{ o } } \) = \(\frac { tan{ 65 }^{ o } }{ tan{ 25 }^{ o } } \) = 1
We are given that sin 3A = cos (A - 26°) ...(1)
Since sin 3A = cos(90° - 3A) we can write (1) as
cos (90° - 3A) = cos (A - 26°)
Since 90° - 3A and A - 26° are both acute angles
90° - 3A = A - 26°
which gives A = 29°
cot 85° + cos 75°
= cot(90° - 5°) + cos(90° - 15°)
= tan 5° + sin 15°
A and B represent points on the bank on opposite sides at the river, so that AB is the width of the river. P is a point on the bridge at a height of 3m i.e., DP = 3 m. We are interested to determine the width at the river which is the length at the side AB of the ΔAPB.
Now, AB = AD + DB
In right ΔAPD,
ㄥA = 30o
So, tan 30° = \(\frac { 1 }{ \sqrt { 3 } } =\frac { 3 }{ AD } \)
i.e., \(\frac { 1 }{ \sqrt { 3 } } =\frac { 3 }{ AD } \) (or) AD = 3\(\sqrt { 3 } \)
Also, in right ΔPBD,
B = 45o
So, BD = PD = 3m
Now, AB = BD + AD
= 3 + 3\(\sqrt { 3 } \) = 3(1+\(\sqrt { 3 } \))m
Therefore, the width at the river is 3(+\(\sqrt { 3 } \))m
\(L H S =\frac{\sin \theta}{1-\cos \theta} \)
\(=\frac{\sin \theta}{1-\cos \theta} \times \frac{1+\cos \theta}{1+\cos \theta} \)
\(=\frac{\sin \theta(1+\cos \theta)}{1-\cos ^{2} \theta} \quad\left[\because 1-\cos ^{2} \theta=\sin ^{2} \theta\right] \)
\(=\frac{1}{\sin \theta}(1+\cos \theta)=\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta} \)
\(=\operatorname{cosec} \theta+\cot \theta=\mathrm{RHS}
\)
\(\text { LHS }=\frac{\tan \theta+\sin \theta}{\tan \theta-\sin \theta}\)
\(=\frac{\frac{\sin \theta}{\cos \theta}+\sin \theta}{\frac{\sin \theta}{\cos \theta}-\sin \theta}\)
\(=\frac{\sin \theta\left(\frac{1}{\cos \theta}+1\right)}{\sin \theta\left(\frac{1}{\cos \theta}-1\right)}=\frac{\sec \theta+1}{\sec \theta-1}=\mathrm{RHS}\)
