Table of Contents
Explanation:
\space\space\space\space\space\space \cos^4 θ - \sin^4 θ = \dfrac{2}{3} \\[4.5 bp]
⇒ \space\space\space\space\space\space (\cos^2 θ)^2 - (\sin^2 θ)^2 = \dfrac{2}{3} \\[4.5 bp]
⇒\space\space\space\space\space\space (\cos^2 θ + \sin^2 θ)(\cos^2 θ - \sin^2 θ) = \dfrac{2}{3} \\[4.5 bp]
⇒\space\space\space\space\space\space \cos^2 θ - \sin^2 θ = \dfrac{2}{3} \\[4.5 bp] \space\space\space\space\space\space(∵ \sin^2 θ + \cos^2 θ = 1) \\[4.5 bp]
⇒\space\space\space\space\space\space 1- \sin^2 θ - \sin^2 θ = \dfrac{2}{3} \\[4.5 bp]\space\space\space\space\space\space(∵ \cos^2 θ = 1- \sin^2 θ) \\[4.5 bp]
⇒\space\space\space\space\space\space 1 - 2 \sin^2 θ =\dfrac{2}{3}
Explanation:
Given : \tan \space A =\dfrac{1}{\sqrt{5}} \\[4.5 bp]
\Rightarrow \space\space\space\space \tan^2{A}=\dfrac{1}{5} \space\space\space\space ...(i) \\[4.5 bp]
\Rightarrow \space\space\space\space
\dfrac{1}{\cot^2A}=\dfrac{1}{5} \\[4.5 bp] \space\space\space\space
\begin{bmatrix}
∵ \space\space\space\space \tan A = \dfrac{1}{\cot A}\end{bmatrix} \\[4.5 bp]
⇒\space\space\space\space \cot^2 A = 5 \space\space\space\space …(ii) \\[5 bp]
\text{Now,} \space \space \dfrac{\cosec^2A - \sec^2A}{\cosec^2A + \sec^2A}=\dfrac{1+\cot^2A-(1+\tan^2A)}{1 + \cot^2A+1 + \tan^2A} \\[4.5 bp]
[∵ \cosec^2 A = 1 + \cot^2 A \space \\
and \space sec^2 A = 1 + \tan^2 A ] \\[5 bp]
=\space\space\space\space
\dfrac{\cot^2A-\tan^2A}{2 + \cot^2A + \tan^2A}=\dfrac{5-\dfrac{1}{5}}{2+5+\dfrac{1}{5}} \\[4.5 bp]
\left[\text{ From (i) and (ii) }\right] \\[4.5 bp]
= \space\space\space\space \dfrac{\dfrac{25-1}{5}}{\dfrac{35+1}{5}}=\dfrac{24}{36} = \dfrac{2}{3} \\[5 bp]
\dfrac{\cosec^2A-\sec^2 A}{\cosec^2A+sec^2 A}=\dfrac{2}{3}
Explanation:
\text{Given, a cos θ – b sin θ = c ...(i)} \\[4.5 bp]
\text{Now, }a^2 + b^2 – c^2 = a^2 + b^2 – (a \cos θ – b \sin θ)^2 \text{\space \space \space \space [From (i)]}\\[4.5 bp]
= a^2 + b^2 – (a^2 cos^2 θ + b^2 \sin^2 θ – 2ab \sin θ cos θ) \\[4.5 bp]
= a^2 + b^2 – [a^2(1 – \sin^2 θ) + b^2(1 – \cos^2 θ) – 2ab \sin θ \cos θ]\\[4.5 bp]
= a^2 + b^2 – [a^2 – a^2 \sin^2 θ + b^2 – b^2 \cos^2 θ – 2ab \sin θ \cos θ]\\[4.5 bp]
= a^2 + b^2 – a^2+ a^2 \sin^2 θ – b^2 + b^2 \cos^2 θ + 2ab \sin θ \cos θ]\\[4.5 bp]
= a^2sin^2 θ + b^2 \cos^2 θ + 2ab \sin θ \cos θ \\[4.5 bp]
= (a \sin θ + b \cos θ)^2\\[4.5 bp]
\text{Thus, }a^2 + b^2 – c^2 = (a \sin θ + b \cos θ)^2\\[4.5 bp]
\text{or }\space \space \space \sqrt{a^2 + b^2 - c^2} = (a\space \sin \spaceθ + b\space \cos \spaceθ)\\[4.5 bp]
\rArr \space \space \space
(a\space \sin \spaceθ + b\space \cos \spaceθ) = \sqrt{a^2 + b^2 - c^2}.
Ans. L.H.S. = R.H.S.
Explanation:
L.H.S. = (\tan\space A + \cosec\space B)^2 – (\cot\space B – \sec\space A)^2 \\[5.5 bp]
= \tan^2 A + \cosec^2 B + 2 \tan A. \cosec\space B – (\cot^2\space B + \sec^2\space A – 2 \cot \space B. \sec\space A) \\[5.5 bp]
= \tan^2\space A + \cosec^2 B + 2 \tan \space A. \cosec\space B – \cot^2\space B – \sec^2\space A + 2 \cot\space B \sec\space A \\[5.5 bp]
= (\cosec^2\space B – \cot^2\space B) – (\sec^2\space A – \tan^2 A) + 2 \tan\space A.\cosec \space B + 2 \cot \space B \sec \space A \\[5.5 bp]
= 1 – 1 + 2 \tan \space A. \cosec \space B + 2 \cot \space B. \sec \space A \\[5.5 bp]
= 2 \tan \space A. \cosec \space B + 2 \cot \space B.\sec \space A \\[5.5 bp]
=2\begin{pmatrix} \dfrac {\sin \space A}{\cos \space A}×\dfrac{1}{\sin B}+\dfrac{\cos \space B}{\sin B}×\dfrac{1}{\cos \space A}\end{pmatrix} \\[5.5 bp]
=2\begin{pmatrix} \dfrac{\sin \space A + \cos \space B}{\cos \space A. \sin \space B}\end{pmatrix}...(i) \\[5.5 bp]
R.H.S. = 2 tan A.cot B.(cosec A + sec B)
= 2 tan A.cot B. cosec A + 2 tan A. cot B.sec B
=2\dfrac{\sin \space A}{\cos\space A}.\dfrac{\cos\space B}{\sin\space A}.\dfrac{1}{\sin\space A}+\dfrac{2 \sin A}{\cos \space A}.\dfrac{\cos B}{\sin\space B}.\dfrac{1}{\cos\space B} \\[5.5 bp] =2\begin{pmatrix} \dfrac{\cos\space B}{\cos\space A.\sin\space B}+\dfrac{sin \space A}{\cos A. \sin B}\end{pmatrix} \\[5.5 bp] =2 \begin{pmatrix} \dfrac{\cos B+ \sin A}{\cos\space A. \sin\space B}\end{pmatrix}\text{ ....(ii)} \\[5.5 bp] \text{ From (i) and (ii), } \\[5.5 bp] \text{L.H.S. = R.H.S. }| Chapter No. | Chapter Name |
|---|---|
| Chapter 1 | Real Numbers |
| Chapter 2 | Polynomials |
| Chapter 3 | Pair of Linear Equations in Two Variable |
| Chapter 4 | Quadratic Equations |
| Chapter 5 | Arithmetic Progressions |
| Chapter 6 | Triangles |
| Chapter 7 | Coordinate Geometry |
| Chapter 8 | Introduction to Trigonometry |
| Chapter 9 | Some Applications of Trigonometry |
| Chapter 10 | Circles |
| Chapter 11 | Areas Related to Circle |
| Chapter 12 | Surface Areas and Volumes |
| Chapter 13 | Statistics |
| Chapter 14 | Probability |
| Chapter Wise Important Questions for CBSE Board Class 10 Maths |
|---|
| Real Numbers |
| Polynomials |
| Pair of Linear Equations in Two Variables |
| Quadratic Equations |
| Arithmetic Progressions |
| Triangles |
| Coordinate Geometry |
| Introduction to Trigonometry |
| Some Applications of Trigonometry |
| Circles |
| Areas Related to Circles |
| Surface Areas and Volumes |
| Statistics |
| Probability |
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