Download App

Trigonometric Identities — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTrigonometric Identities5 Marks
Question
If $(\text{cosec }\theta+\sin\theta)=\text{a}^3$ and $(\sec\theta-\cos\theta)=\text{b}^3,$ prove that $\big(\text{a}^2\text{b}^2\big)\big(\text{a}^2+\text{b}^2\big)=1.$
✓

Answer

We have $(\text{cosec }\theta-\sin\theta)=\text{a}^3$ and $(\sec\theta-\cos\theta)=\text{n}$
$\Rightarrow\text{a}^3=\Big(\frac{1}{\sin\theta}-\sin\theta\Big)$
$\Rightarrow\text{a}^3=\frac{\big(1-\sin^2\theta\big)}{\sin\theta}$
$=\frac{\cos^2\theta}{\sin\theta}$
$\therefore\ \text{a}=\frac{\cos^{\frac23}\theta}{\sin^\frac13\theta}$
Again, $(\sec\theta-\cos\theta)=\text{b}^3$
$\Rightarrow\text{b}^3=\Big(\frac{1}{\cos\theta}-\cos\theta\Big)$
$=\frac{\big(1-\cos^2\theta\big)}{\cos\theta}$
$=\frac{\sin^2\theta}{\cos\theta}$
$\therefore\ \text{b}=\frac{\sin^\frac{2}{3}\theta}{\cos^\frac13\theta}$
Now, $\text{LHS}=\text{a}^2\text{b}^2\big(\text{a}^2+\text{b}^2\big)$
$=\text{a}^4\text{b}+\text{a}^2\text{b}^4$
$=\text{a}^3\big(\text{ab}^2\big)+\big(\text{a}^2\text{b}\big)\text{b}^3$
$=\frac{\cos^3\theta}{\sin\theta}\times\Bigg[\frac{\cos^\frac23\theta}{\sin^\frac13\theta}\times\frac{\sin^\frac43\theta}{\cos^\frac23\theta}\Bigg]\\ \ +\frac{\cos^3\theta}{\sin\theta}\times\Bigg[\frac{\cos^\frac23\theta}{\sin^\frac13\theta}\times\frac{\sin^\frac43\theta}{\cos^\frac23\theta}\Bigg]\times\frac{\sin^2\theta}{\cos\theta}$
$=\frac{\cos^2\theta}{\sin\theta}\times\sin\theta+\cos\theta\times\frac{\sin^2\theta}{\cos\theta}$
$=\cos^2\theta+\sin^2\theta=1$
$=\text{RHS}$
Hence proved.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Question" from Trigonometric Identities→Trigonometric Identities — all question types→Maths STD 10 question papers→Maharashtra Board STD 10 question papers→Maharashtra Board question paper generator→

Similar questions

A tent is made in the form of a frustum of a cone surmounted by another cone. The diameters of the base and the top of the frustum are 20m and 6m, respectively, and the height is 24m. If the height of the tent is 28m and the radius of the conical part is equal to the radius of the top of the frustum, find the quantity of canvas required. $\Big[\text{Take }\pi=\frac{22}{7}.\Big]$
→
Construct a $\triangle\text{ABC},$ with $BC = 7cm,$ $\angle\text{B}=60^\circ$ and AB = 6cm. Construct another triangle whose sides are $\frac34\text{times}$ the corresponding sides of $\triangle\text{ABC}.$
→
A bucket is in the form of a frustum of a cone with a capacity of $12308.8\ cm^3 $ of water. The radii of the top and bottom circular ends are $20\ cm$ and $12\ cm$, respectively. Find the height of the bucket. $[$Use $\pi = 3.14]$
→
If $\tan\text{A}=\text{n}\tan\text{B}$ and $\sin\text{A}=\text{m}\sin\text{B},$ prove that $\cos^2\text{A}=\frac{\big(\text{m}^2-1\big)}{\text{n}^2-1}.$
→
A copper wire of diameter 6mm is evenly wrapped on a cylinder of length 18cm and diameter 49cm to cover its whole surface. Find the length and the volume of the wire. If the density of the copper be 8.8g per cu-cm, then find the weight of the wire.
→
Find the arithmetic mean of each of the following frequency distributions using step-deviation method:
Class
500-520
520-540
540-560
560-580
580-600
600-620
Frequency
14
9
5
4
3
5
→
In an AP, the first term is $-4$, the last term is $29$ and the sum of all its terms is $150$. Find the common difference.
→
Prove that the angle between the two tangent drawn from an external point to a circle is supplenentary to the angle subtended by the line segments joining the points of contact at the centre.
→
Solve the following systems of equations by using the method of cross multiplication:
$\frac{\text{x}}{6}+\frac{\text{y}}{15}=4,$
$\frac{\text{x}}{3}-\frac{\text{y}}{15}=\frac{\text{19}}{4}$
→
Draw a circle of radius 3cm. From a point P, 7cm away from the centre of the circle, draw two tangents to the circle. Also, measure the lengths of the tangents.
→