Question 14 Marks
Prove the following trigonometric identities.
$\frac{\sin\text{A}}{\sec\text{A}+\tan\text{A}-1}+\frac{\cos\text{A}}{\text{cosec A}+\cot\text{A}-1}=1$
Answer$\text{L.H.S}=\frac{\sin\text{A}}{\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}-1}+\frac{\cos\text{A}}{\frac{1}{\sin\text{A}}+\frac{\cos\text{A}}{\sin\text{A}}-1}$
$=\frac{\sin\text{A}}{\frac{1+\sin\text{A}-\cos\text{A}}{\cos\text{A}}}+\frac{\cos\text{A}}{\frac{1+\cos\text{A}-\sin\text{A}}{\sin\text{A}}}$
$=\frac{\sin\text{A}\cos\text{A}}{1+\sin\text{A}-\cos\text{A}}+\frac{\sin\text{A}\cos\text{A}}{1+\cos\text{A}-\sin\text{A}}$
$=\sin\text{A}\cos\text{A}\Big[\frac{1}{1+\sin\text{A}-\cos\text{A}}+\frac{1}{1+\cos\text{A}-\sin\text{A}}\Big]$
$=\sin\text{A}\cos\text{A}\Big[\frac{1+\cos\text{A}-\sin\text{A}+1+\sin\text{A}-\cos\text{A}}{(1+\sin\text{A}-\cos\text{A})(1+\cos\text{A}-\sin\text{A})}\Big]$
$=\sin\text{A}\cos\text{A}\Big[\frac{2}{1+\cos\text{A}-\sin\text{A}+\sin\text{A}+\sin\text{A}\cos\text{A}-\sin^2\text{A}-\cos\text{A}-\cos^2\text{A}+\cos\text{A}\sin\text{A}}\Big]$
$=\sin\text{A}\cos\text{A}\Big[\frac{2}{1-\sin^2\text{A}-\cos^2\text{A}+2\sin\text{A}\cos\text{A}}\Big]$
$=\sin\text{A}\cos\text{A}\Big[\frac{2}{1-(\sin^2\text{A}+\cos^2\text{A})+2\sin\text{A}\cos\text{A}}\Big]$
$=\sin\text{A}\cos\text{A}\Big[\frac{2}{1-1+2\sin\text{A}\cos\text{A}}\Big](\because \sin^2\text{A}+\cos^2\text{A}=1)$
$=\sin\text{A}\times\cos\text{A}\times\frac{2}{2\sin\text{A}\cos\text{A}}$
$=1=\text{R.H.S}$
$\text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 24 Marks
Prove the following trigonometric identities.
$\frac{\cot^2\text{A}(\sec\text{A}-1)}{1+\sin\text{A}}=\sec^2\text{A}\Big(\frac{1-\sin\text{A}}{1+\sin\text{A}}\Big)$
AnswerWe have to prove $\frac{\cot^2\text{A}(\sec\text{A}-1)}{1+\sin\text{A}}=\sec^2\text{A}\Big(\frac{1-\sin\text{A}}{1+\sin\text{A}}\Big)$
We know that, $\sin^2\text{A}+\cos^2\text{A}=1$
So,
$\text{L.H.S}=\frac{\cot^2\text{A}(\sec\text{A}-1)}{1+\sin\text{A}}=\sec^2\text{A}\Big(\frac{1-\sin\text{A}}{1+\sin\text{A}}\Big)$
$=\frac{\frac{\cos^2\text{A}}{\sin^\text{2}\text{A}}\big(\frac{1}{\cos\text{A}}-1\Big)}{1+\sin\text{A}}$
$=\frac{\frac{\cos^2\text{A}}{\sin^2\text{A}}\frac{1-\cos\text{A}}{\cos\text{A}}}{1+\sin\text{A}}$
$=\frac{\cos\text{A}(1-\cos\text{A})}{\sin^2\text{A}(1+\sin\text{A})}$
$=\frac{\cos\text{A}(1-\cos\text{A})}{(1-\cos^2\text{A})(1+\sin\text{A})}$
$=\frac{\cos\text{A}(1-\cos\text{A})}{(1-\cos\text{A})(1+\cos\text{A})(1+\sin\text{A})}$
$=\frac{\cos\text{A}}{(1+\cos\text{A})(1+\sin\text{A})}$
$=\frac{\frac{1}{\sec\text{A}}}{\Big(1+\frac{1}{\sec\text{A}}\Big)(1+\sin\text{A})}$
$=\frac{\frac{1}{\sec\text{A}}}{\Big(\frac{\sec\text{A}+1}{\sec\text{A}}\Big)(1+\sin\text{A})}$
$=\frac{1}{(\sec\text{A}+1)(1+\sin\text{A})}$
Multiplying both the numerator ans denominator by $(1-\sin\text{A})$ we have
$=\frac{(1-\sin\text{A})}{(\sec\text{A}+1)(1+\sin\text{A})(1-\sin\text{A})}$
$=\frac{1-\sin\text{A}}{(\sec\text{A}+1)\cos^2\text{A}}$
$=\frac{(1-\sin\text{A})}{(\sec\text{A}+1)\cos^2\text{A}}$
$=\sec^2\text{A}\frac{(1-\sin\text{A})}{(\sec\text{A}+1)}$
$=\sec^2\text{A}\Big(\frac{1-\sin\text{A}}{1+\sec\text{A}}\Big)=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 34 Marks
Prove the following trigonometric identities.
$\frac{\tan^2\text{A}}{1+\tan^2\text{A}}+\frac{\cot^2\text{A}}{1+\cot^2\text{A}}=1$
AnswerIn the given question, we need to prove $\frac{\tan^2\text{A}}{1+\tan^2\text{A}}+\frac{\cot^2\text{A}}{1+\cot^2\text{A}}=1$
Here, we will first solve the L.H.S.
Now using $\tan\theta=\frac{\sin\theta}{\cos\theta}\text{ and }\cot\theta=\frac{\cos\theta}{\sin\theta},$ we get
$\text{L.H.S}=\frac{\tan^2\text{A}}{1+\tan^2\text{A}}+\frac{\cot^2\text{A}}{1+\cot^2\text{A}}$
$=\frac{\Big(\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)}{\Big(1+\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)}+\frac{\Big(\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)}{\Big(1+\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)}$
$=\frac{\Big(\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)}{\Big(\frac{\cos^2\text{A}+\sin^2\text{A}}{\cos^2\text{A}}\Big)}+\frac{\Big(\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)}{\Big(\frac{\sin^2\text{A}+\cos^2\text{A}}{\sin^2\text{A}}\Big)}$
$=\frac{\Big(\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)}{\Big(\frac{1}{\cos^2\text{A}}\Big)}+\frac{\Big(\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)}{\Big(\frac{1}{\sin^2\text{A}}\Big)} \ (\text{Using }\sin^2\theta+\cos^2\theta=1)$
On further solving by taking the reciprocal of the denominator, we get,
$=\frac{\Big(\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)}{\Big(\frac{1}{\cos^2\text{A}}\Big)}+\frac{\Big(\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)}{\Big(\frac{1}{\sin^2\text{A}}\Big)}$
$=\Big(\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)\Big(\frac{\cos^2\text{A}}{1}\Big)+\Big(\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)\Big(\frac{\sin^2\text{A}}{1}\Big)$
$=\sin^2\text{A}+\cos^2\text{A}\ (\text{using} \sin^2\theta+\cos^2\theta)$
$=1=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 44 Marks
Prove the following trigonometric identities.
If $\text{T}_\text{n}=\sin^\text{n}\theta+\cos^\text{n}\theta,$ porve that $\frac{\text{T}_3-\text{T}_5}{\text{T}_1}=\frac{\text{T}_5-\text{T}_7}{\text{T}_3}.$
AnswerIn the given question, we are given $\text{T}_\text{n}=\sin^\text{n}\theta+\cos^\text{n}\theta$
We need to prove $\frac{\text{T}_3-\text{T}_5}{\text{T}_1}=\frac{\text{T}_5-\text{T}_7}{\text{T}_3}$
Here L.H.S is
$\text{L.H.S}=\frac{\text{T}_3-\text{T}_5}{\text{T}_1}=\frac{(\sin^3\theta+\cos^3\theta)-(\sin^5\theta+\cos^5\theta)}{(\sin\theta+\cos\theta)}$
Now, solving the L.H.S, we get
$\frac{(\sin^3\theta+\cos^3\theta)-(\sin^5\theta+\cos^5\theta)}{(\sin\theta+\cos\theta)}=\frac{\sin^3\theta-\sin^3\theta+\cos^3\theta-\cos^3\theta}{\sin\theta+\cos\theta}$
$=\frac{\sin^3\theta(1-\sin^2\theta)+\cos^3\theta(1-\cos^2\theta)}{\sin\theta+\cos\theta}$
Further using the property $\sin^2\theta+\cos^2\theta=1,$ we get
$\cos^2\theta=1-\sin^2\theta$
$\sin^2\theta=1-\cos^2\theta$
$=\frac{\sin^3\theta(1-\sin^2\theta)+\cos^3\theta(1-\cos^2\theta)}{\sin\theta+\cos\theta}=\frac{\sin^3\theta\cos^2\theta+\cos^3\theta\sin^2\theta}{\sin\theta+\cos\theta}$
$=\frac{\sin^2\theta\cos^2\theta(\sin\theta+\cos\theta)}{\sin\theta+\cos\theta}$
$=\sin^2\theta\cos^2\theta$
Now, solving the R.H.S, we get
$\text{R.H.S}=\frac{\text{T}_5-\text{T}_7}{\text{T}_3}=\frac{(\sin^5\theta+\cos^5\theta)-(\sin^7\theta+\cos^7\theta)}{(\sin^3\theta+\cos^3\theta)}$
So,
$\frac{(\sin^5\theta+\cos^5\theta)-(\sin^7\theta+\cos^7\theta)}{(\sin^3\theta+\cos^3\theta)}=\frac{\sin^5\theta-\sin^7\theta+\cos^5\theta-\cos^7\theta}{\sin^3\theta+\cos^3\theta}$
$=\frac{\sin^5\theta(1-\sin^2\theta)+\cos^5\theta(1-\cos^2\theta)}{\sin^3\theta+\cos^3\theta}$
Furhter using the property $\sin^2\theta+\cos^2\theta=1,$ we get,
$\text{cos}^2\theta=1-\sin^2\theta$
$\sin^2\theta=1-\cos^2\theta$
So,
$=\frac{\sin^5\theta(1-\sin^2\theta)+\cos^5\theta(1-\cos^2\theta)}{\sin^3\theta+\cos^3\theta}=\frac{\sin^5\theta\cos^2\theta+\cos^5\theta\sin^2\theta}{\sin^3\theta+\cos^3\theta}$
$=\frac{\sin^2\theta\cos^2\theta(\sin^3\theta+\cos^3\theta)}{\sin^3\theta+\cos^3\theta}$
$=\sin^2\theta\cos^2\theta$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 54 Marks
Prove the following trigonometric identities.
$(\text{cosec }\theta-\sec\theta)(\cot\theta-\tan\theta)=(\text{cosec }\theta+\sec\theta)(\sec\theta\text{ cosec }\theta-2)$
AnswerWe have to prove $(\text{cosec }\theta-\sec\theta)(\cot\theta-\tan\theta)=(\text{cosec }\theta+\sec\theta)(\sec\theta\text{ cosec }\theta-2)$
We know that, $\sin^2\theta+\cos^2\theta=1$
Consider the L.H.S
$\text{L.H.S}=(\text{cosec }\theta-\sec\theta)(\cot\theta-\tan\theta)$
$=\Big(\frac{1}{\sin\theta}-\frac{1}{\cos\theta}\Big)\Big(\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}\Big)$
$=\Big(\frac{\cos\theta-\sin\theta}{\sin\theta\cos\theta}\Big)\Big(\frac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}\Big)$
$=\frac{(\cos\theta-\sin\theta)}{\sin\theta\cos\theta}\frac{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)}{\sin\theta\cos\theta}$
$=\frac{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)^2}{\sin^2\theta\cos^2\theta}$
Now, consider the R.H.S
$\text{R.H.S}=(\text{cosec }\theta+\sec\theta)(\sec\theta\text{ cosec }\theta-2)$
$=\Big(\frac{1}{\sin\theta}+\frac{1}{\cos\theta}\Big)\Big(\frac{1}{\cos\theta}\frac{1}{\sin\theta}-2\Big)$
$=\Big(\frac{\cos\theta+\sin\theta}{\sin\theta\cos\theta}\Big)\Big(\frac{1-2\sin\theta\cos\theta}{\sin\theta\cos\theta}\Big)$
$=\frac{(\cos\theta+\sin\theta)}{\sin\theta\cos\theta}\frac{(\cos^2\theta+\sin^2\theta-2\cos\theta\sin\theta)}{\sin\theta\cos\theta}$
$=\frac{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)^2}{\sin^2\theta\cos^2\theta}$
$\therefore \text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 64 Marks
Prove the following trigonometric identities.
$\frac{\cos\theta}{1-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$
Answer$\text{L.H.S}=\frac{\cos\theta}{1-\sin\theta}$
Taking rationalisation
$=\frac{\cos\theta(1+\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}$
$=\frac{\cos\theta(1+\sin\theta)}{1-\sin^2\theta}$
$=\frac{\cos\theta(1+\sin\theta)}{\cos^2\theta}$
$=\frac{1+\sin\theta}{\cos\theta}=\text{R.H.S}$
View full question & answer→Question 74 Marks
Prove the following trigonometric identities.
$\frac{1}{\sec\text{A}-1}+\frac{1}{\sec\text{A}+1}=2\text{cosec A}\cot\text{A}$
AnswerWe need to prove $\frac{1}{\sec\text{A}-1}+\frac{1}{\sec\text{A}+1}=2\text{cosec A}\cot\text{A}$
Solving the L.H.S, we get
$\text{L.H.S}=\frac{1}{\sec\text{A}-1}+\frac{1}{\sec\text{A}+1}=\frac{\sec\text{A}+1+\sec \text{A}-1}{(\sec\text{A}-1)(\sec\text{A}+1)}$
$=\frac{2\sec\text{A}}{\sec^2\text{A}-1}$
Further using the property $1+\tan^2\theta=\sec^2\theta,$ we get
So,
$\frac{2\sec\text{A}}{\sec^2\text{A}-1}=\frac{2\sec\text{A}}{\tan^2\text{A}}$
$=\frac{2\Big(\frac{1}{\cos\text{A}}\Big)}{\frac{\sin^2\text{A}}{\cos^2\text{A}}}$
$=2\frac{1}{\cos\text{A}}\times\frac{\cos^2\text{A}}{\sin^2\text{A}}$
$=2\Big(\frac{\cos\text{A}}{\sin\text{A}}\Big)\times\frac{1}{\sin\text{A}}$
$=2\text{cosec A}\cot\text{A}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 84 Marks
Prove the following trigonometric identities.
If $\text{a}\cos\theta+\text{b}\sin\theta=\text{m and a}\sin\theta-\text{b}\cos\theta=\text{n},$ prove that $a^2 + b^2 = m^2 + n^2$.
AnswerGiven,
$\text{a}\cos\theta+\text{b}\sin\theta=\text{m},$
$\text{a}\sin\theta-\text{b}\cos\theta=\text{n}$
We have to prove $a^2 + b^2 = m^2 + n^2$
We know that, $\sin^2\theta+\cos^2\theta=1$
Now, squaring and adding the two equations, we get
$(\text{a}\cos\theta+\text{b}\sin\theta)^2+(\text{a}\sin\theta-\text{b}\cos\theta)^2=\text{m}^2+\text{n}^2$
$\Rightarrow \big(\text{a}^2\cos^2\theta+2\text{ab}\sin\theta\cos\theta+\text{b}^2\sin^2\theta\big)$
$+\big(\text{a}^2\sin^2\theta-2\text{ab}\sin\theta\cos\theta+\text{b}^2\cos^2\theta\big)=2$
$\Rightarrow\ \text{a}^2(\cos^2\theta+\sin^2\theta)+\text{b}^2(\sin^2\theta+\cos^2\theta)=\text{m}^2+\text{n}^2$
$\Rightarrow\ \text{a}^2+\text{b}^2=\text{m}^2+\text{n}^2$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 94 Marks
Prove the following trigonometric identities.
$\frac{1-\cos\text{A}}{1+\cos\text{A}}=(\cot\text{A}-\text{cosec A})^2$
Answer$\text{L.H.S}=\frac{(1-\cos\text{A})}{(1+\cos\text{A})}\times\frac{(1-\cos\text{A})}{(1-\cos\text{A})}$
$=\frac{(1-\cos\text{A})^2}{(1-\cos^2\text{A})}$
$=\frac{(1-\cos\text{A})^2}{\sin^2\text{A}}$
$=\Big(\frac{1}{\sin\text{A}}-\frac{\cos\text{A}}{\sin\text{A}}\Big)^2$
$=(\text{cosec A}-\cot\text{A})^2$
$=(\cot\text{A}-\text{cosec A})^2$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 104 Marks
Prove the following trigonometric identities.
If $\text{x}=\text{a}\sec\theta\cos\phi,\text{y}=\text{b}\sec\theta\sin\phi\text{ and z}=\text{c}\tan\theta,$ show that $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}-\frac{\text{z}^2}{\text{c}^2}=1.$
Answer$\text{x}=\text{a}\sec\theta\cos\phi$
$\Rightarrow\ \frac{\text{x}}{\text{a}}=\sec\theta\cos\phi\ .....\text{i}$
$\text{y}=\text{b}\sec\theta\sin\phi$
$\Rightarrow \frac{\text{y}}{\text{b}}=\sec\theta\sin\theta \ .....(\text{ii})$
$\Rightarrow\ \frac{\text{z}}{\text{c}}=\tan\theta\ .....\text{(iii)}$
We have to prove that $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}-\frac{\text{z}^2}{\text{c}^2}=1$
Squaring the above equations and then subtracting the third from the sum of the first two, we have
$\Big(\frac{\text{x}}{\text{a}}\Big)^2+\Big(\frac{\text{y}}{\text{b}}\Big)^2-\Big(\frac{\text{z}}{\text{c}}\Big)^2=(\sec\theta\cos\phi)^2+(\sec\theta\sin\theta)-(\tan\theta)^2$
$\Rightarrow\ \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=\sec^2\theta\cdot\cos^2\phi+\sec^2\theta\sin^2\phi-\tan^2\theta$
$\Rightarrow\ \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=(\sec^2\cos^2\phi+\sec^2\theta\sin^2\phi)-\tan^2\theta$
$\Rightarrow\ \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=\sec^2\theta(\cos^2\phi+\sin^2\phi)-\tan^2\theta$
$\Rightarrow\ \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=\sec^2\theta(1)-\tan^2\theta$
$\Rightarrow\ \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=\sec^2\theta-\tan^2\theta$
$\Rightarrow\ \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=1$
Hence proved.
View full question & answer→Question 114 Marks
Prove the following trigonometric identities.
If $\sin\theta+2\cos\theta=1$ prove that $2\sin\theta-\cos\theta=2.$
AnswerIf is given that,
$\text{L.H.S}=\sin\theta+2\cos\theta=1$
$\Rightarrow\ 2\cos\theta=1-\sin\theta\ .....(\text{i})$
Squarign both sides, we get
$\Rightarrow\ (2\cos\theta)^2=(1-\sin\theta)^2$
$\Rightarrow\ 4\cos^2\theta=1+\sin^2\theta-2\sin\theta$
$\Rightarrow\ 4\cos^2\theta=1+\sin^2\theta-2\sin\theta-1+1$
$\Rightarrow\ 4\cos^2\theta=2-2\sin\theta-(1-\sin^2\theta)$
$\Rightarrow\ 4\cos^2\theta=2-2\sin\theta-\cos^2\theta$
$\Rightarrow\ 5\cos^2\theta=2(1-\sin\theta)$
$\Rightarrow\ 5\cos^2\theta=4\cos\theta \ [\text{Using (i)}]$
$\Rightarrow\ \cos\theta(5\cos\theta-4)=0$
$\Rightarrow\ \cos\theta=0,\cos\theta=\frac{4}{5}$
Putting $\cos\theta=\frac{4}{5}\text{ in }\sin\theta+2\cos\theta=1,$ we get
$\Rightarrow\ \sin\theta=1-2\cos\theta=1-2\Big(\frac{4}{5}\Big)=-\frac{3}{5}$
This is not possible.
Putting $\cos\theta=0\text{ in }\sin\theta+2\cos\theta=1,$ we get
$\sin\theta=1$
Thus, the value of $2\sin\theta-\cos\theta=2(1)-0=2.=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 124 Marks
Prove the following trigonometric identities.
$(\sec\text{A}-\text{cosec A})(1+\tan\text{A}+\cot\text{A})=\tan\text{A}\sec\text{A}-\cot\text{A}\text{ cosec A}$
Answer$\text{L.H.S}=(\sec\text{A}-\text{cosec A})(1+\tan\text{A}+\cot\text{A})$
$=\Big[\frac{1}{\cos\text{A}}-\frac{1}{\sin\text{A}}\Big]\Big[1+\frac{\sin\text{A}}{\cos\text{A}}+\frac{\cos\text{A}}{\sin\text{A}}\Big]$
$=\Big[\frac{\sin\text{A}-\cos\text{A}}{\sin\text{A}\cos\text{A}}\Big]\Big[\frac{\cos\text{A}\sin\text{A}+\sin^2\text{A}+\cos^2\text{A}}{\sin\text{A}\cos\text{A}}\Big]$
$=\frac{(\sin\text{A}-\cos\text{A})(\sin^2\text{A}+\cos\text{A}\sin\text{A}+\cos^2\text{A})}{\sin^2\text{A}\cos^2\text{A}}$
$=\frac{(\sin^3\text{A}-\cos^3\text{A})}{\sin^2\text{A}\cos^2\text{A}} \big[\therefore\ (\text{a}-\text{b})(\text{a}^2+\text{ab})+\text{b}=(\text{a}^3-\text{b}^2)\big]$
$=\text{R.H.S}=\tan\text{A}\sec\text{A}-\cot\text{A}\text{ cosec A}$
$=\frac{\sin\text{A}}{\cos\text{A}}\times\frac{1}{\cos\text{A}}-\frac{\cos\text{A}}{\sin\text{A}}\times\frac{1}{\sin\text{A}}$
$=\frac{\sin\text{A}}{\cos^2\text{A}}-\frac{\cos\text{A}}{\sin^2\text{A}}$
$=\frac{\sin^3\text{A}-\cos^3\text{A}}{\sin^2\text{A}\cos^2\text{A}}$
$\text{L.H.S}=\text{R.H.S}$
Hence proved.
View full question & answer→Question 134 Marks
Prove the following trigonometric identities.
$\Big(\frac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}\Big)^2=\frac{1-\cos\theta}{1+\cos\theta}$
AnswerIn the given question, we need to prove $\Big(\frac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}\Big)^2=\frac{1-\cos\theta}{1+\cos\theta}$
Taking $\sin\theta$ common from the numerator and the denominator if the L.H.S, we get
$\text{L.H.S}=\Big(\frac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}\Big)^2=\Big(\frac{(\sin\theta)(\text{cosec }\theta+1-\cot\theta)}{(\sin\theta)(\text{cosec }\theta+1+\cot\theta)}\Big)^2$
$=\Big(\frac{1+\text{cosec }\theta-\cot\theta}{1+\text{cosec }\theta+\cot\theta}\Big)^2$
Now, using the property $1+\cot^2\theta=\text{cosec}^2\theta,$ we get
$\Big(\frac{1+\text{cosec }\theta-\cot\theta}{1+\text{cosec }\theta+\cot\theta}\Big)^2=\bigg(\frac{(\text{cosec}^2\theta-\cot^2\theta)+\text{cosec }\theta-\cot\theta}{1+\text{cosec }\theta+\cot\theta}\bigg)^2$
Using $a^2 - b^2 = (a + b)(a - b)$, we get
$\bigg(\frac{(\text{cosec}^2\theta-\cot^2\theta)+\text{cosec }\theta-\cot\theta}{1+\text{cosec }\theta+\cot\theta}\bigg)^2$
$=\bigg(\frac{(\text{cosec }\theta+\cot\theta)(\text{cosec}-\cot\theta)+(\text{cosec }\theta-\cot)}{1+\text{cosec}\theta+\cot\theta}\bigg)^2$
Taking $\text{cosec }\theta-\cot\theta$ common from the numerator, we get
$=\bigg(\frac{(\text{cosec }\theta+\cot\theta)(\text{cosec}-\cot\theta)+(\text{cosec }\theta-\cot)}{1+\text{cosec}\theta+\cot\theta}\bigg)^2$
$=\bigg(\frac{(\text{cosec }\theta-\cot\theta)(\text{cosec }\theta+\cot\theta+1)}{1+\text{cosec }\theta+\cot\theta}\bigg)^2$
$=(\text{cosec }\theta-\cot\theta)^2$
Using $\cot\theta=\frac{\cos\theta}{\sin\theta}\text{ and cosec }\theta=\frac{1}{\sin\theta},$ we get
$(\text{cosec }\theta-\cot\theta)^2=\Big(\frac{1}{\sin\theta}-\frac{\cos\theta}{\sin\theta}\Big)^2$
$=\Big(\frac{1-\cos\theta}{\sin\theta}\Big)^2$
$=\frac{(1-\cos\theta)^2}{\sin^2\theta}$
Now, using the property $\sin^2\theta+\cos^2\theta=1,$ we get
$\frac{(1-\cos\theta)^2}{\sin^2\theta}=\frac{(1-\cos\theta)^2}{1-\cos^2\theta}$
$=\frac{(1-\cos\theta)^2}{(1+\cos\theta)(1-\cos\theta)}$
$=\frac{1-\cos\theta}{1+\cos\theta}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 144 Marks
Prove the following trigonometric identities.
$(1+\cot\text{A}+\tan\text{A})(\sin\text{A}-\cos\text{A})=\frac{\sec\text{A}}{\text{cosec}^2\text{A}}-\frac{\text{cosec A}}{\sec^2\text{A}}=\sin\text{A}\tan\text{A}-\cot\text{A}\cos\text{A}$
Answer$\text{L.H.S}=(1+\cot\text{A}+\tan\text{A})(\sin\text{A}-\cos\text{A})$
$=\sin\text{A}-\cos\text{A}+\cot\text{A}\sin\text{A}-\cot\text{A}\cos\text{A}\\\ +\sin\text{A}\tan\text{A}-\tan\text{A}\cos\text{A}$
$$$=\sin\text{A}-\cos\text{A}+\frac{\cos\text{A}}{\sin\text{A}}\times\sin\text{A}-\cot\text{A}\cos\text{A}\\\ +\sin\text{A}\tan\text{A}-\frac{\sin\text{A}}{\cos\text{A}}\times\cos\text{A}$
$=\sin\text{A}-\cos\text{A}+\cos\text{A}-\cot\text{A}\cos\text{A}+\sin\text{A}\tan\text{A}-\sin\text{A}$
$=\sin\text{A}\cos\text{A}\cos\text{A}\cot\text{A}$
Solving:
$\frac{\sec\text{A}}{\text{cosec}^2\text{A}}-\frac{\text{cosec A}}{\sec^2\text{A}}$
$\frac{\frac{1}{\cos\text{A}}}{\frac{1}{\sin^2\text{A}}}-\frac{\frac{1}{\sin\text{A}}}{\frac{1}{\cos^2\text{A}}}$
$\frac{\sin^2\text{A}}{\cos\text{A}}-\frac{\cos^2\text{A}}{\sin\text{A}}$
$\frac{\sin^3\text{A}-\cos^3\text{A}}{\sin\text{A}\cos\text{A}}$
$=\sin\text{A}\times\frac{\sin\text{A}}{\cos\text{A}}-\cos\text{A}\times\frac{\cos\text{A}}{\sin\text{A}}$
$=\sin\text{A}\tan\text{A}-\cos\text{A}\cot\text{A}=\text{R.H.S}$
$\text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 154 Marks
Prove the following trigonometric identities.
If $\frac{\text{x}}{\text{a}}\cos\theta+\frac{\text{y}}{\text{b}}\sin\theta=1\text{ and }\frac{\text{y}}{\text{a}}\sin\theta-\frac{\text{y}}{\text{b}}\cos\theta=1,$ prove that $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=2.$
AnswerGiven that,
$\frac{\text{x}}{\text{a}}\cos\theta+\frac{\text{y}}{\text{b}}\sin\theta=1\ .....\text{(i)}$
$\frac{\text{x}}{\text{a}}\sin\theta-\frac{\text{y}}{\text{b}}\cos\theta=1\ .....\text{(ii)}$
We have to prove $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=2$
We know that, $\sin^2\theta+\cos^2\theta=1$
Squaring and then adding the above two equations, we have
$\text{L.H.S}=\Big(\frac{\text{x}}{\text{a}}\cos\theta+\frac{\text{y}}{\text{b}}\sin\theta\Big)^2+\Big(\frac{\text{x}}{\text{a}}\sin\theta-\frac{\text{y}}{\text{b}}\cos\theta\Big)^2=1+1$
$\Rightarrow \Big(\frac{\text{x}^2}{\text{a}^2}\cos^2\theta+2\frac{\text{xy}}{\text{ab}}\sin\theta\cos\theta+\frac{\text{y}^2}{\text{b}^2}\sin^2\theta\Big)$
$=\Big(\frac{\text{x}^2}{\text{a}^2}\sin^2\theta-2\frac{\text{xy}}{\text{ab}}\sin\theta\cos\theta+\frac{\text{y}^2}{\text{b}^2}\cos^2\theta\Big)=2$
$\Rightarrow\ \frac{\text{x}^2}{\text{a}^2}(\cos^2\theta+\sin^2\theta)+\frac{\text{y}^2}{\text{b}^2}(\sin^2\theta+\cos^2\theta)=2$
$\Rightarrow \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=2=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 164 Marks
Prove the following trigonometric identities.
$\frac{\tan^3\theta}{1+\tan^2\theta}+\frac{\cot^3\theta}{1+\cot^2\theta}=\sec\theta\text{cosec}\theta-2\sin\theta\cos\theta$
AnswerIn the given question, we need to prove $\frac{\tan^3\theta}{1+\tan\theta}+\frac{\cot^3\theta}{1+\cot^2\theta}=\sec\theta\text{cosec}\theta-2\sin\theta\cos\theta$
Using the property $1+\tan^2\theta=\sec^2\theta\text{ and }1+\cot^3\theta=\text{cosec}^2\theta,$ we get
$\text{L.H.S}=\frac{\tan^3\theta}{1+\tan^2\theta}+\frac{\cot^3\theta}{1+\cot^2\theta}=\frac{\tan^3\theta}{\sec^2\theta}+\frac{\cot^3\theta}{\text{cosec}^2\theta}$
$=\frac{\Big(\frac{\sin^2\theta}{\cos^3\theta}\Big)}{\Big(\frac{1}{\cos^2\theta}\Big)}+\frac{\Big(\frac{\cos^3\theta}{\sin^3\theta}\Big)}{\Big(\frac{1}{\sin^2\theta}\Big)}$
Taking the reciprocal of the denominator, we get
$=\frac{\Big(\frac{\sin^2\theta}{\cos^3\theta}\Big)}{\Big(\frac{1}{\cos^2\theta}\Big)}+\frac{\Big(\frac{\cos^3\theta}{\sin^3\theta}\Big)}{\Big(\frac{1}{\sin^2\theta}\Big)}$
$=\Big(\frac{\sin^3\theta}{\cos^3\theta}\times\frac{\cos^2\theta}{1}\Big)+\Big(\frac{\cos^3\theta}{\sin^3\theta}\times\frac{\sin^2\theta}{1}\Big)$
$=\frac{\sin^3\theta}{\cos\theta}+\frac{\cos^3\theta}{\sin\theta}$
$=\frac{\sin^4\theta+\cos^4\theta}{\cos\theta\sin\theta}$
$=\frac{(\sin^2\theta)^2+(\cos^2\theta)^2}{\cos\theta\sin\theta}$
Further, using the identity $a^2 + b^2 = (a + b)^2 - 2ab$, we get
$=\frac{(\sin^2\theta)^2+(\cos^2\theta)^2}{\cos\theta\sin\theta}=\frac{(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta}{\cos\theta\sin\theta}$
$=\frac{1}{\cos\theta\sin\theta}-\frac{2\sin^2\theta\cos^2\theta}{\cos\theta\sin\theta}(\text{using}\sin^2\theta+\cos^2\theta = 1)$
$=\frac{1}{\cos\theta\sin\theta}-2\sin\theta\cos\theta$
$=\sec\theta\text{ cosec }\theta-2\sin\theta\cos\theta=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 174 Marks
Prove the following trigonometric identities.
If $\cos\text{A}+\cos^2\text{A}=1,$ prove that $\sin^{12}\theta+\sin^{10}\theta+3\sin^8\theta+\sin^6\theta+2\sin^4\theta+2\sin^2\theta-2=1.$
AnswerGiven,
$\cos\theta+\cos^2\theta=1$
We have to prove $\sin^{12}\theta+\sin^{10}\theta+3\sin^8\theta+\sin^6\theta+2\sin^4\theta+2\sin^2\theta-2=1.$
From the given equation, we have
$\cos\theta+\cos^2\theta=1$
$\Rightarrow \cos\theta=1-\cos^2\theta$
$\Rightarrow \cos\theta=\sin^2\theta$
$\Rightarrow \sin^2\theta=\cos\theta$
Therefore, we have
$\text{L.H.S}=\sin^{12}\theta+\sin^{10}\theta+3\sin^8\theta+\sin^6\theta+2\sin^4\theta+2\sin^2\theta-2$
$=(\sin^{12}\theta+\sin^{10}\theta+3\sin^8\theta+\sin^6\theta)+2(\sin^4\theta+\sin^2\theta)-2$
$=\{(\sin^4\theta)^3+3(\sin^4\theta)^2\sin^2\theta+3\sin^4\theta(\sin^2\theta)^2+(\sin^2\theta)^3\}$
$=2(\sin^4+\sin^2\theta)-2$
$=(\sin^4\theta+\sin^2\theta)^3+2(\sin^4\theta+\sin^2\theta)-2$
$=(\cos^2\theta+\cos\theta)^3+2(\cos^2\theta+\cos\theta)-2$
$=(1)^3+2(1)-2$
$=1=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 184 Marks
Prove the following trigonometric identities.
$\Big(\frac{1}{\sec^2\theta-\cos^2\theta}+\frac{1}{\text{cosec}^2\theta-\sin^2\theta}\Big)\sin^2\theta\cos^2\theta=\frac{1-\sin^2\theta\cos^2\theta}{2+\sin^2\theta\cos^2\theta}$
Answer$\text{L.H.S}=\Big(\frac{1}{\sec^2\theta-\cos^2\theta}+\frac{1}{\text{cosec}^2\theta-\sin^2\theta}\Big)\sin^2\theta\cos^2\theta$
$\Rightarrow \bigg[\frac{1}{\frac{1}{\cos^2\theta}-\cos^2\theta}+\frac{1}{\text{cosec}^2\theta-\sin^2\theta}\bigg]\sin^2\theta\cos^2\theta$
$=\Bigg[\frac{1}{\frac{1-\cos^4\theta}{\cos^2\theta}}+\frac{1}{\frac{1-\sin^4\theta}{\sin^2\theta}}\Bigg]\sin^2\theta\cos^2\theta$
$=\Big[\frac{\cos^2\theta}{1-\cos^4\theta}+\frac{\sin^2\theta}{1-\sin^4\theta}\Big]\sin^2\theta\cos^2\theta$
$=\Big[\frac{\cos^2\theta}{\cos^2\theta+\sin^2\theta-\cos^4\theta}+\frac{\sin^2}{\cos^2\theta+\sin^2\theta-\sin^4\theta}\Big]\sin^2\theta\cos^2\theta$
$=\Big[\frac{\cos^2\theta}{\cos^2\theta(1-\cos^2\theta)+\sin^2\theta}+\frac{\sin^2}{\sin^2\theta(1-\sin^2\theta)+\cos^2\theta}\Big]\sin^2\theta\cos^2\theta$
$=\Big[\frac{\cos^2\theta}{\cos^2\theta\sin^2\theta+\sin^2\theta}+\frac{\sin^2\theta}{\sin^2\theta\cos^2\theta+\cos^2\theta}\Big]\sin^2\theta\cos^2\theta$
$=\bigg[\frac{\cos^2\theta}{\sin^2\theta(\cos^2\theta+1)}+\frac{\sin^2\theta}{\cos^2\theta(\sin^2\theta+1)}\bigg]\sin^2\theta\cos^2\theta$
$=\bigg[\frac{\cos^4\theta(1+\sin^2\theta)+\sin^4\theta(1+\cos^2\theta)}{\sin^2\theta\cos^2\theta(1+\cos^2\theta)(1+\sin^2\theta)}\bigg]\sin^2\theta\cos^2\theta$
$=\frac{\cos^4\theta(1+\sin^2\theta)+\sin^4\theta(1+\cos^2\theta)}{(1+\cos^2\theta)(1+\sin^2\theta)}$
$=\frac{\cos^4\theta+\cos^4\theta\sin^2\theta+\sin^4\theta+\sin^4\theta\cos^2\theta}{1+\sin^2\theta+\cos^2\theta+\cos^2\theta\sin^2\theta}$
$=\frac{1-2\sin^2\theta\cos^2\theta+\sin^2\theta\cos^2\theta(\cos^2\theta+\sin^2\theta)}{1+1+\cos^2\theta\sin^2\theta}$
$(\because\ \cos^2\theta+\sin^2\theta=1)$
$=\frac{1-\sin^2\theta\cos^2\theta}{2+\sin^2\theta\cos^2\theta}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 194 Marks
If $\sin\theta+\cos\theta=\text{x},$ prove that $\sin^6\theta+\cos^6\theta=\frac{4-3(\text{x}^2-1)^2}{4}.$
AnswerGiven,
$\sin\theta+\cos\theta=\text{x}$
Squaring the given equation, we have
$(\sin\theta+\cos\theta)^2=\text{x}^2$
$\Rightarrow\ \sin^2+2\sin\theta\cos\theta+\cos^2\theta=\text{x}^2$
$\Rightarrow\ (\sin^2\theta+\cos^2\theta)+2\sin\theta\cos\theta=\text{x}^2$
$\Rightarrow\ 1+2\sin\theta\cos\theta=\text{x}^2$
$\Rightarrow\ 2\sin\theta\cos\theta=\text{x}^2-1$
$\Rightarrow \sin\theta\cos\theta=\frac{\text{x}^2-1}{2}$
Squaring the last equation, we have
$(\sin\theta\cos\theta)^2=\frac{(\text{x}^2-1)^2}{4}$
$\Rightarrow \sin^2\theta\cos^2\theta=\frac{(\text{x}^2-1)^2}{4}$
Therefore, we have
$\sin^6\theta+\cos^6\theta=(\sin^2\theta)^3+(\cos^2\theta)^3$
$=(\sin^2\theta+\cos^2\theta)^3-3\sin^2\theta\cos^2\theta(\sin^2\theta+\cos^2\theta)$
$=(1)^3-3\frac{(\text{x}^2-1)^2}{4}(1)$
$=1-3\frac{(\text{x}^2-1)^2}{4}$
$=\frac{4-3(\text{x}^2-1)^2}{4}$
Hence proved.
View full question & answer→Question 204 Marks
Prove the following trigonometric identities.
$(1+\tan^2\text{A})+\Big(1+\frac{1}{\tan^2\text{A}}\Big)=\frac{1}{\sin^2\text{A}-\sin^4\text{A}}$
AnswerWe need to prove $(1+\tan^2\text{A})+\Big(1+\frac{1}{\tan^2\text{A}}\Big)=\frac{1}{\sin^2\text{A}-\sin^4\text{A}}$
Using the property $1+\tan^2\theta=\sec^2\theta,$ we get
$\text{L.H.S}=(1+\tan^2\text{A})+\Big(1+\frac{1}{\tan^2\text{A}}\Big)=\sec^2\text{A}+\Big(\frac{\tan^2\text{A}+1}{\tan^2\text{A}}\Big)$
$=\sec^2\text{A}+\Big(\frac{\sec^2\text{A}}{\tan^2\text{A}}\Big)$
Now, using $\sec\theta=\frac{1}{\cos\theta}\text{ and }\tan\theta=\frac{\sin\theta}{\cos\theta},$ we get
$\sec^2\text{A}+\Big(\frac{\sec^2\text{A}}{\tan^2\text{A}}\Big)=\frac{1}{\cos^2\text{A}}+\Bigg(\frac{\frac{1}{\cos^2\text{A}}}{\frac{\sin^2\text{A}}{\cos^2\text{A}}}\Bigg)$
$=\frac{1}{\cos^2\text{A}}+\Big(\frac{1}{\cos^2\text{A}}\times\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)$
$=\frac{1}{\cos^2\text{A}}+\frac{1}{\sin^2\text{A}}$
$=\frac{\sin^2\text{A}+\cos^2\text{A}}{\cos^2\text{A}(\sin^2\text{A})}$
Further, using the peroperty, $\sin^2\theta+\cos^2\theta=1,$ we get
$\frac{\sin^2\text{A}+\cos^2\text{A}}{\cos^2\text{A}(\sin^2\text{A})}=\frac{1}{\cos^2\text{A}(\sin^2\text{A})}$
$=\frac{1}{(1-\sin^2\text{A})(\sin^2\text{A})}\ (\text{using}\cos^2\theta=1-\sin^2\theta)$
$=\frac{1}{\sin^2\text{A}-\sin^4\text{A}}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 214 Marks
Prove the following trigonometric identities.
If $\text{cosec }\theta-\sin\theta=\text{a}^3,\sec\theta-\cos\theta=\text{b}^3,$ prove that $a^2b^2 (a^2 + b^2) = 1$.
Answer$\text{cosec }\theta-\sin\theta=\text{a}^3$
$\frac{1}{\sin\theta}-\sin\theta=\text{a}^3$
$\frac{1-\sin^2}{\sin\theta}=\text{a}^3$
$\frac{\cos^2\theta}{\sin\theta}=\text{a}^3$
$\text{a}=\frac{\cos^{\frac{2}{3}}\theta}{\sin^{\frac{1}{3}}\theta}$
$\Rightarrow\ \text{a}^2=\frac{\cos\frac{4}{3}\theta}{\sin\frac{2}{3}\theta}$
$\sec\theta-\cos\theta=\text{B}^3$
$\frac{1}{\cos\theta}-\cos\theta=\text{b}^3$
$\frac{1-\cos^2\theta}{\cos\theta}=\text{b}^3$
$\frac{\sin^2\theta}{\cos\theta}=\text{b}^3$
$\text{b}=\frac{\sin\frac{2}{3}\theta}{\cos\frac{1}{3}\theta}$
Now, $\text{L.H.S}=\text{a}^2\text{b}^2(\text{a}^2+\text{b}^2)$
$=\frac{\cos^{\frac{4}{3}}\theta}{\sin^{\frac{2}{3}}\theta}\times\frac{\sin^\frac{4}{3}\theta}{\cos^\frac{2}{3}\theta}\bigg(\frac{\cos^{\frac{4}{3}}\theta}{\sin^{\frac{2}{3}\theta}}+\frac{\sin^{\frac{4}{3}}\theta}{\cos^{\frac{2}{3}}\theta}\bigg)$
$=\cos^{\frac{4}{3}-\frac{2}{3}}\theta\sin^{\frac{4-2}{3}}\bigg(\frac{\cos^{\frac{4}{3}}\theta}{\sin^{\frac{2}{3}\theta}}+\frac{\sin^{\frac{4}{3}}\theta}{\cos^{\frac{2}{3}}\theta}\bigg)$
$=\cos^{\frac{2}{3}}\theta\sin^\frac{2}{3}\bigg(\frac{1}{\sin^{\frac{2}{3}}\theta\cos^{\frac{2}{3}}\theta}\bigg) \big(\because \cos^2\theta+\sin^2\theta=1\big)$
$=1=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 224 Marks
Prove the following trigonometric identities.
Given that: $(1+\cos\alpha)(1+\cos\beta)(1+\cos\gamma)=(1-\cos\alpha)(1-\cos\alpha)(1-\cos\beta)(1-\cos\gamma)$
Show that one of the values of each member of this equality is $\sin\alpha\sin\beta\sin\gamma.$
AnswerL.H.S
We know that $1+\cos\theta=1+\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}=2\cos^2\frac{\theta}{2}$
$\therefore\ \Rightarrow 2\cos^2\frac{\alpha}{2}\times2\cos^2\frac{\beta}{2}\times2\cos^2\frac{\gamma}{2}\ .....\text{(i)}$
Multiply (i) with $\sin\alpha\sin\beta\sin\gamma$ and divide it with samw we get
$\frac{8\cos^2\frac{\alpha}{2}\cos^2\frac{\beta}{2}\cos^2\frac{\gamma}{2}}{\sin\alpha\sin\beta\sin\gamma}\times\sin\alpha\sin\beta\sin\gamma$
$\Rightarrow \frac{2\cos^2\frac{\alpha}{2}\cos^2\frac{\beta}{2}\cos^2\frac{\gamma}{2}\times\sin\alpha\sin\beta\sin\gamma}{\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}}$
$\Rightarrow \sin\alpha\sin\beta\sin\gamma \times {\cot\frac{\alpha}{2}\cot\frac{\beta}{2}\cot\frac{\gamma}{2}}$
$\text{R.H.S} \ (1-\cos\alpha)(1-\cos\beta)(1-\cos\gamma)$
We know that $1-\cos\theta=1-\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}=2\cos^2\frac{\theta}{2}$
$\Rightarrow\ 2\sin^2\frac{\alpha}{2}2\times\sin^2\frac{\beta}{2}\times2\sin^2\frac{\gamma}{2}$
Multiply and divide by $\sin\alpha\sin\beta\sin\gamma$ we get
$\Rightarrow\ \frac{2\sin^2\frac{\alpha}{2}2\sin^2\frac{\beta}{2}2\sin^2\frac{\gamma}{2}\times\sin\alpha\sin\beta\sin\gamma}{\sin\alpha\sin\beta\sin\gamma}$
$\Rightarrow\ \frac{2\sin^2\frac{\alpha}{2}2\sin^2\frac{\beta}{2}2\sin^2\frac{\gamma}{2}\times\sin\alpha\sin\beta\sin\gamma}{2\sin\frac{\alpha}{2}\cos\frac{\beta}{2}\sin\frac{\beta}{2}\cos\frac{\beta}{2}2\sin\frac{\gamma}{2}\cos\frac{\gamma}{2}}$
$\Rightarrow\ \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\tan\frac{\gamma}{2}\sin\alpha\sin\beta\sin\gamma$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Hence $\sin\alpha\sin\beta\sin\gamma$ is the member of equality.
View full question & answer→