Question 14 Marks
Prove the following identities:
$\frac{\tan\theta}{(\sec\theta-1)}+\frac{\tan\theta}{(\sec\theta+1)}=2\text{cosec}\theta$
Answer$\text{LHS}=\frac{\tan\theta}{(\sec\theta-1)}+\frac{\tan\theta}{(\sec\theta+1)}$
$=\frac{\frac{\sin\theta}{\cos\theta}}{\big(\frac{1}{\cos\theta}-1\big)}+\frac{\frac{\sin\theta}{\cos\theta}}{\big(\frac{1+\cos\theta}{\cos\theta}\big)}$
$=\frac{\sin\theta}{1-\cos\theta}+\frac{\sin\theta}{1+\cos\theta}$
$=\frac{\sin\theta(1+\cos\theta)+\sin\theta(1-\cos\theta)}{1-\cos^2\theta}$
$=\frac{\sin\theta+\sin\theta\cos\theta+\sin\theta-\sin\theta\text{ cosec}\theta}{\sin^2\theta}$
$=\frac{2\sin\theta}{\sin^2\theta}=\frac{2}{\sin\theta}$
$=2\text{ cosec}\theta$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 24 Marks
Prove the following identities:
$(1+\tan\theta+\cot\theta)(\sin\theta-\cos\theta)=\Big(\frac{\sec\theta}{\text{cosec}^2\theta}-\frac{\text{cosec}\theta}{\sec^2\theta}\Big)$
Answer$\text{LHS}=(1+\tan\theta+\cot\theta)(\sin\theta-\cos\theta)$
$=\Big(1+\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}\Big)(\sin\theta-\cos\theta)$
$=\Big(\frac{\cos\theta\sin\theta+\sin^2\theta+\cos^2\theta}{\cos\theta\sin\theta}\Big)(\sin\theta-\cos\theta)$
$=\frac{(\cos\theta\sin\theta+1)}{\cos\theta\sin\theta}(\sin\theta-\cos\theta)$
$\text{RHS}=\Big(\frac{\sec\theta}{\text{cosec}^2\theta}-\frac{\text{cosec }\theta}{\sec^2\theta}\Big)$
$=\Bigg(\frac{\frac{1}{\cos\theta}-\frac{1}{\sin\theta}}{\frac{1}{\sin^2\theta}\frac{1}{\cos^2\theta}}\Bigg)$
$=\Big(\frac{\sin^2\theta}{\cos\theta}-\frac{\cos^2\theta}{\sin\theta}\Big)=\frac{\sin^3\theta-\cos^3\theta}{\cos\theta\sin\theta}$
$=\frac{(\sin\theta-\cos\theta)\big(\sin^2\theta+\cos^2\theta+\cos\theta\sin\theta\big)}{\cos\theta\sin\theta}$ $\Big[\because\text{a}^3-\text{b}^3=(\text{a}-\text{b})\big(\text{a}^2+\text{ab}+\text{b}^2\big)\Big]$
$=\frac{(\sin\theta-\cos\theta)(1+\cos\theta\sin\theta)}{\cos\theta\sin\theta}$
$\therefore\ \text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 34 Marks
Prove the following identities:
If $1+\sin^2\theta=3\sin\theta\cos\theta$ then prove that $\tan\theta=1$ or $\frac12.$
Answer$1+\sin^2\theta=3\sin\theta\cos\theta$
Write 1 $=\sin^2\theta+\cos^2\theta$
$2\sin^2\theta+\cos^2\theta=3\sin\theta\cos\theta$
Divide both side by $\cos^2\theta$
$2\tan^2\theta+1=3\tan\theta$
$2\tan^2\theta-3\tan\theta+1=0$
$\tan\theta=\frac{3\pm\sqrt{9-8}}{2(2)}$
$=\frac{3\pm1}{4}$
$=1$ or $\frac12.$
View full question & answer→Question 44 Marks
Prove the following identities:
$\frac{1-\tan^2\theta}{1+\tan^2\theta}=\big(\cos^2\theta-\sin^2\theta\big)$
Answer$\text{LHS}=\frac{1-\tan^2\theta}{1+\tan^2\theta}$
$=\frac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}$
$=\frac{\Big(\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}\Big)}{\Big(\frac{\cos^2\theta+\sin^\theta}{\cos^2\theta}\Big)}=\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}$
$=\frac{\big(\cos^2\theta-\sin^2\theta\big)}{1}=\big(\cos^2\theta-\sin^2\theta\big)$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 54 Marks
Prove the following identities:
$\frac{\sec\theta-\tan\theta}{\sec\theta+\tan\theta}=\frac{\sin^2\theta}{(1+\cos\theta)^2}$
Answer$\text{LHS}=\frac{\sec\theta-\tan\theta}{\sec\theta+\tan\theta}$
$=\frac{\sec\theta-1}{\sec\theta+1}=\frac{\big(\frac{1}{\cos\theta}-1\big)}{\big(\frac{1}{\cos\theta}+1\big)}$
$=\frac{1-\cos\theta}{1+\cos\theta}$
$=\frac{(1+\cos\theta)}{(1+\cos\theta)}\times\frac{(1+\cos\theta)}{(1+\cos\theta)}$
$=\frac{1-\cos^2\theta}{(1+\cos\theta)^2}$
$=\frac{\sin^2\theta}{(1+\cos\theta)^2}$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 64 Marks
Prove the following identities:
$\Bigg\{\frac{1}{\big(\sec^2\theta-\cos^2\theta\big)}+\frac{1}{\big(\text{cosec}^2\theta-\sin^2\theta\big)}\Bigg\}\big(\sin^2\theta\cos^2\theta\big)$
$=\frac{1-\sin^2\theta\cos^2\theta}{2+\sin^2\theta\cos^2\theta}$
Answer$\text{LHS}=\Bigg\{\frac{1}{\big(\sec^2\theta-\cos^2\theta\big)}+\frac{1}{\big(\text{cosec}^2\theta-\sin^2\theta\big)}\Bigg\}\big(\sin^2\theta\cos^2\theta\big)$
$=\Bigg\{\frac{1}{\frac{1}{\cos^2\theta}-\cos^2\theta}+\frac{1}{\frac{1}{\sin^2\theta}-\sin^2\theta}\Bigg\}\times\sin^2\theta\cos^2\theta$
$=\Bigg\{\frac{\sin^2\theta\cos^2\times\cos^2\theta}{\big(1-\cos^4\theta\big)}+\frac{\sin^4\theta\cos^2\theta\sin^2\theta}{1-\sin^4\theta}\Bigg\}$
$=\Bigg\{\frac{\sin^2\theta\times\cos^4\theta}{\big(1-\cos^2\theta\big)\big(1-\cos^2\theta\big)}+\frac{\sin^4\theta\cos^2\theta}{1-\sin^2\theta\big(1+\sin^2\theta\big)}\Bigg\}$
$=\Bigg\{\frac{\cos^4\theta}{\big(1+\cos^2\theta\big)}+\frac{\sin^4\theta}{1+\sin^2\theta}\Bigg\}$
$=\frac{\cos^4\theta+\cos^4\theta\sin^2\theta+\sin^4\theta+\sin^4\theta\cos^2\theta}{\big(1+\cos^2\theta\big)\big(1+\sin^2\theta\big)}$
$=\frac{\cos^4\theta+\sin^4\theta+\cos^2\theta\sin^2\theta\big(\cos^2\theta+\sin^2\theta\big)}{1+\sin^2\theta+\cos^2\theta+\cos^2\theta+\sin^2\theta}$ $\Big[\because\text{a}^2+\text{b}^2=(\text{a}+\text{b})^2-2\text{ab}\Big]$
$=\frac{1-\cos^2\theta\sin^2\theta}{2+\cos^2\theta\sin^2\theta}$
$=\text{R.H.S}$
$\therefore\ \text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 74 Marks
Prove the following identities:
$\frac{\tan\theta}{(1-\cot\theta)}+\frac{\cot\theta}{(1-\tan\theta)}=(1+\sec\theta\text{ cosec }\theta)$
Answer$\text{L.H.S.}=\frac{\tan\theta}{(1-\cot\theta)}+\frac{\cot\theta}{(1-\tan\theta)}$
$=\frac{\frac{\sin\theta}{\cos\theta}}{1-\frac{\cos\theta}{\sin\theta}}+\frac{\frac{\cos\theta}{\sin\theta}}{1-\frac{\sin\theta}{\cos\theta}}$ $\Big[\because\tan\theta=\frac{\sin\theta}{\cos\theta},\cot\theta=\frac{\cos\theta}{\sin\theta}\Big]$
$=\frac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)}+\frac{\cos^2\theta}{\sin\theta(\cos\theta-\sin\theta)}$
$=\frac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)}+\frac{\cos^2\theta}{\sin\theta(\cos\theta-\sin\theta)}$
$=\frac{\sin^3\theta-\cos^3\theta}{\sin\theta\cos\theta(\sin\theta-\cos\theta)}$
$=\frac{(\sin\theta-\cos\theta)\big(\sin^2\theta+\cos^2\theta+\sin\theta\cos\theta\big)}{(\sin\theta-\cos\theta)\sin\theta\cos\theta}$ $\Big[\because\text{a}^3-\text{b}^3=(\text{a}-\text{b})\big(\text{a}^2+\text{ab}+\text{b}^2\big)\Big]$
$=\frac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta}$
$=\frac{1}{\sin\theta\cos\theta}+1$
$=1+\sec\theta\text{ cosec }\theta$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 84 Marks
Prove the following identities:
$\frac{\sin\theta+1-\cos\theta}{\cos\theta-1+\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$
Answer$\text{LHS}=\frac{\sin\theta+1-\cos\theta}{\cos\theta-1+\sin\theta}$
On dividing numerator and denominator of LHS $\cos^\theta,$
We, get
$\text{LHS}=\frac{\tan\theta+\sec\theta-1}{1-\sec\theta+\tan\theta}$
$=\frac{(\tan\theta+\sec\theta)+\big(\sec^2\theta-\tan^2\theta\big)}{1-\sec\theta+\tan\theta}$
$\big($Writing 1 $=\sec^2\theta-\tan^2\theta\big)$
$=\frac{(\tan\theta+\sec\theta)+(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{(1-\sec\theta+\tan\theta)}$
$=\frac{(\tan\theta+\sec\theta)(1-\sec\theta+\tan\theta)}{(1-\sec\theta+\tan\theta)}$
$=\tan\theta+\sec\theta=\frac{\sin\theta}{\cos\theta}+\frac{1}{\cos\theta}$
$=\frac{\sin\theta+1}{\cos\theta}$
$=\text{R.H.S.}$
$\therefore\ \text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 94 Marks
Prove the following identities:
$\frac{\tan\theta}{\big(1+\tan^2\theta\big)}+\frac{\cot\theta}{\big(1+\cot^2\theta\big)}=\sin\theta\cos\theta$
Answer$\text{L.H.S.}=\frac{\tan\theta}{\big(1+\tan^2\theta\big)}+\frac{\cot\theta}{\big(1+\cot^2\theta\big)}$
$=\frac{\tan\theta}{\big(\sec^2\theta\big)^2}+\frac{\cot\theta}{\big(\text{cosec}^2\theta\big)^2}$
$=\frac{\sin\theta}{\cos\theta}\times\frac{1}{\sec^4\theta}+\frac{\cos\theta}{\sin\theta}\times\frac{1}{\text{cosec}^4\theta}$
$=\frac{\sin\theta}{\cos\theta}\times\cos^4\theta+\frac{\cos\theta}{\sin\theta}\times\sin^4\theta$
$=\sin\theta\cos^3\theta+\cos\theta\sin^3\theta$
$=\sin\theta\cos\theta\big(\cos^2\theta+\sin^2\theta\big)$
$=\sin\theta\cos\theta$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 104 Marks
Prove the following identities:
$\big(1+\tan^2\theta\big)\big(1+\cot^2\theta\big)=\frac{1}{\big(\sin^2\theta-\sin^4\theta\big)}$
Answer$\text{L.H.S.}=\big(1+\tan^2\theta\big)\big(1+\cot^2\theta\big)$
$=\sec^2\theta\text{ cosec}^2\theta$
$=\frac{1}{\sin^2\theta\cos^2\theta}=\frac{1}{\sin^2\theta\big(1-\sin^2\theta\big)}$
$=\frac{1}{\sin^2\theta-\sin^4\theta}$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 114 Marks
Prove the following identities:
$\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}+\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}=2\text{cosec}\theta$
Answer$\text{LHS}=\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}+\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$
$=\sqrt{\frac{1+\cos\theta}{1-\cos\theta}\times\frac{1+\cos\theta}{1+\cos\theta}}\\ \ +\sqrt{\frac{1-\cos\theta}{1+\cos\theta}\times\frac{1-\cos\theta}{1-\cos\theta}}$
$=\sqrt{\frac{(1+\cos\theta)^2}{1-\cos^2\theta}}+=\sqrt{\frac{(1-\cos\theta)^2}{1-\cos^2\theta}}$
$=\sqrt{\frac{(1+\cos\theta)^2}{\sin^2\theta}}+\sqrt{\frac{(1-\cos\theta)^2}{\sin^2\theta}}$
$=\frac{1+\cos\theta}{\sin\theta}+\frac{1-\cos\theta}{\sin\theta}$
$=\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}+=\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}$
$=\text{cosec }\theta+\text{cosec }\theta$
$=2\text{cosec }\theta$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 124 Marks
Prove the following identities:
$\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}=(\sec\theta+\tan\theta)$
Answer$\text{LHS}=\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}$
$\sqrt{\frac{1+\sin\theta}{1-\sin\theta}\times\frac{1+\sin\theta}{1+\sin\theta}}$
$\sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}}$
$\sqrt{\frac{(1+\sin\theta)^2}{\cos^2\theta}}$
$=\frac{1+\sin\theta}{\cos\theta}$
$=\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}$
$=\sec\theta+\tan\theta$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 134 Marks
Prove the following identities:
$\text{cosec}^4\theta-\text{cosec}^2\theta=\cot^4\theta+\cot^2\theta$
Answer$\text{LHS}=\text{cosec}^4\theta-\text{cosec}^2\theta$
$=\text{cosec}^2\theta\big(\text{cosec}^2\theta-1\big)$
$=\big(1+\cot^2\theta\big)\cot^2\theta$
$=\cot^2\theta+\cot^4\theta$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 144 Marks
Prove the following identities:
$\frac{\cos\theta\text{ cosec }\theta-\sin\theta\sec\theta}{\cos\theta+\sin\theta}=\text{cosec }\theta-\sec\theta$
Answer$\text{LHS}=\frac{\cos\theta\text{ cosec }\theta-\sin\theta\sec\theta}{\cos\theta+\sin\theta}$
$=\frac{\cos\theta\times\frac{1}{\sin\theta}-\sin\theta\times\frac{1}{\cos\theta}}{\cos\theta+\sin\theta}=\frac{\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}}{(\cos\theta+\sin\theta)}$
$=\frac{\cos^2\theta-\sin^2\theta}{\sin^2\theta\cos\theta(\cos\theta+\sin\theta)}$
$=\frac{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)}{\sin\theta\cos\theta(\cos\theta+\sin\theta)}$
$=\frac{\cos\theta}{\sin\theta\cos\theta}-\frac{\sin\theta}{\sin\theta\cos\theta}$
$=\text{cosec }\theta-\sec\theta$
$=\text{R.H.S.}$
$\therefore\ \text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 154 Marks
Prove the following identities:
$\frac{\tan^2\theta}{\big(1+\tan^2\theta\big)}+\frac{\cot^2\theta}{\big(1+\cot^2\theta\big)}=1$
Answer$\text{L.H.S.}=\frac{\tan^2\theta}{\big(1+\tan^2\theta\big)}+\frac{\cot^2\theta}{\big(1+\cot^2\theta\big)}$
$=\frac{\tan^2\theta}{\sec^2\theta}+\frac{\cot^2\theta}{\text{cosec}^2\theta}$ $\Big[\because\big(1+\tan^2\theta\big)=\sec^2\theta$ and $\big(1+\cot^2\theta\big)=\text{cosec}^2\theta\Big]$
$=\frac{\frac{\sin^2\theta}{\cos^2\theta}}{\frac{1}{\cos^2\theta}}+\frac{\frac{\cos^2\theta}{\sin^2\theta}}{\frac{1}{\sin^2\theta}}$
$=\sin^2\theta+\cos^2\theta=1$
$=\text{R.H.S.}$
Hence, LHS = RHS.
View full question & answer→Question 164 Marks
Prove the following identities:
$\sin^2\theta+\cos^4\theta=\cos^2\theta+\sin^4\theta$
Answer$\text{LHS}=\sin^2\theta+\cos^4\theta$
$=\sin^2\theta+\big(\cos^2\theta\big)^2$
$=\sin^2\theta+\big(1-\sin^2\theta\big)^2$
$=\sin^2\theta+1-2\sin^2\theta+\sin^4\theta$
$=1-\sin^2\theta+\sin^4\theta$
$=\cos^2\theta+\sin^4\theta$
$=\text{R.H.S}$
Hence, $\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 174 Marks
Prove the following identities:
$\frac{\cos^3\theta+\sin^3\theta}{\cos\theta+\sin\theta}+\frac{\cos^3\theta-\sin^3\theta}{\cos\theta-\sin\theta}=2$
Answer$\text{LHS}=\frac{\cos^3\theta+\sin^3\theta}{\cos\theta+\sin\theta}+\frac{\cos^3\theta-\sin^3\theta}{\cos\theta-\sin\theta}$
$=\frac{(\cos\theta+\sin\theta)\big(\cos^2\theta-\cos\theta\sin\theta+\sin^2\theta\big)}{\cos\theta+\sin\theta}$
$=\frac{(\cos\theta-\sin\theta)\big(\cos^2\theta+\cos\theta\sin\theta+\sin^2\theta\big)}{\cos\theta-\sin\theta}$
$=\cos^2\theta-\cos\theta\times\sin\theta+\sin^2\theta\\\ \ \ +\cos^2\theta+\cos\theta\sin\theta+\sin^2\theta$
$=2\big[\cos^2\theta+\sin^2\theta\big]=2$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 184 Marks
Prove the following identities:
$\frac{\sec\theta-\tan\theta}{\sec\theta+\tan\theta}=\frac{\cos^2\theta}{(1+\sin\theta)^2}$
Answer$\text{LHS}=\frac{\sec\theta-\tan\theta}{\sec\theta+\tan\theta}$
$=\frac{\sec\theta-\tan\theta}{\sec\theta+\tan\theta}=\frac{\big(\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta}\big)}{\big(\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}\big)}$
$=\frac{(1-\sin\theta)}{(1+\sin\theta)}$
$=\frac{(1+\sin\theta)}{(1+\sin\theta)}\times\frac{(1+\sin\theta)}{(1+\sin\theta)}$
$=\frac{1-\sin^2\theta}{(1+\sin\theta)^2}$
$=\frac{\cos^2\theta}{(1+\sin\theta)^2}$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 194 Marks
Prove the following identities:
$\frac{\text{cosec }\theta+\cot\theta}{\text{cosec }\theta-\cot\theta}=(\text{cosec }\theta+\cot\theta)^2$
$=1+2\cot^2\theta+2\text{ cosec }\theta\cot\theta$
Answer$\text{LHS}=\frac{\text{cosec }\theta+\cot\theta}{\text{cosec }\theta-\cot\theta}\times\frac{(\text{cosec}\theta+\cot\theta)}{(\text{cosec}\theta+\cot\theta)}$
$=\frac{(\text{cosec}\theta+\cot\theta)^2}{\big(\text{cosec}^2\theta-\cot^2\theta\big)}=(\text{cosec}\theta+\cot\theta)^2$
Further,
$(\text{cosec}\theta+\cot\theta)^2$
$=\text{cosec}^2\theta+\cot^2\theta+2\text{cosec }\theta\cot\theta$
$=1+\cot^2\theta+\cot^2\theta+2\text{cosec }\theta\cot\theta$
$=1+2\cot^2+2\text{cosec }\theta\cot\theta$
$\therefore\text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 204 Marks
Prove the following identities:
$\frac{\sin\theta}{(\cot\theta+\text{cosec }\theta)}-\frac{\sin\theta}{(\cot\theta-\text{cosec }\theta)}=2$
Answer$\text{LHS}=\frac{\sin\theta}{(\cot\theta+\text{cosec }\theta)}-\frac{\sin\theta}{(\cot\theta-\text{cosec }\theta)}$
$=\frac{\sin\theta}{\text{cosec }\theta+\cot\theta}+\frac{\sin\theta}{\text{cosec }\theta-\cot\theta}$
$=\frac{\sin\theta(\text{cosec }\theta-\cot\theta)-\sin\theta(\text{cosec }\theta+\cot\theta)}{\text{cosec}^2\theta-\cot^2\theta}$ $\big[\because1+\cot^2\theta=\text{cosec}^2\theta$ and $\text{cosec}^2\theta-\cot^2\theta=1\big]$
$=2\sin\theta\text{cosec }\theta=2\sin\theta\times\frac{1}{\sin\theta}$
$=2$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 214 Marks
Prove the following identities:
$\frac{\sin\theta}{(1+\cos\theta)}+\frac{(1+\cos^2\theta)}{\sin\theta}=2\text{cosec}\theta$
Answer$\text{L.H.S.}=\frac{\sin\theta}{(1+\cos\theta)}+\frac{(1+\cos^2\theta)}{\sin\theta}$
$=\frac{\sin\theta(1-\cos\theta)}{(1+\cos\theta)(1-\cos\theta)}+\frac{1+\cos\theta}{\sin\theta}$
$=\frac{\sin\theta(1-\cos\theta)}{1-\cos^2\theta}+\frac{1+\cos\theta}{\sin\theta}$
$=\frac{\sin\theta(1-\cos\theta)}{\sin\theta}+\frac{1+\cos\theta}{\sin\theta}$
$=\frac{(1-\cos\theta)}{\sin\theta}+\frac{1+\cos\theta}{\sin\theta}$
$=\frac{1-\cos\theta+1+\cos\theta}{\sin\theta}$
$=\frac{2}{\sin\theta}$
$=2\text{cosec}\theta$
$=\text{R.H.S.}$
View full question & answer→Question 224 Marks
Prove the following identities:
$\frac{\big(1+\tan^2\theta\big)\cot\theta}{\text{cosec}^2\theta}=\tan\theta$
Answer$\text{L.H.S.}=\frac{\big(1+\tan^2\theta\big)\cot\theta}{\text{cosec}^2\theta}=\frac{\sec^2\theta\cot\theta}{\text{cosec}^2\theta}$
$=\frac{1}{\cos^2\theta}\times\frac{\cos\theta}{\sin\theta}\times\sin^2\theta=\frac{\sin\theta}{\cos\theta}$ $\Big[\because\big(1+\tan^2\theta\big)=\sec^2\theta\Big]$
$=\tan\theta$
$=\text{R.H.S.}$
Hence, LHS = RHS.
View full question & answer→Question 234 Marks
Prove the following identities:
Show that none of the following is an identity:
$\cos^2\theta+\cos\theta=1$
Answer$\cos^2\theta+\cos\theta=1$
$\text{L.H.S.}=\cos^2\theta+\cos\theta$
$=1-\sin^2\theta+\cos\theta$
$=1-\big(\sin^2\theta-\cos\theta\big)$
Since $\text{L.H.S}\neq\text{R.H.S},$ this is not an identity.
View full question & answer→Question 244 Marks
Prove the following identities:
$\sin^6\theta+\cos^6\theta=1-3\sin^2\theta\cos^2\theta$
AnswerTo prove $\sin^6\theta+\cos^6\theta=1-3\sin^2\theta\cos^2\theta$
We know, $\text{a}^3+\text{b}^3=(\text{a}+\text{b})^3-3\text{ab}(\text{a}+\text{b})$
Put $\text{a}=\sin^2\theta,\text{b}=\cos^2\theta$
$\therefore\sin^6\theta+\cos^6\theta=\big(\sin^2\theta+\cos^2\theta\big)^3\\ \ \ -3\sin^2\theta\cos^2\theta\times\big(\sin^2\theta+\cos^2\theta\big)$
$=1-3\sin^2\theta\cos^2\theta$
$=\text{R.H.S}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 254 Marks
Prove the following identities:
$\sec\theta(1-\sin\theta)(\sec\theta+\tan\theta)=0$
Answer$\text{L.H.S.}=\sec\theta(1-\sin\theta)(\sec\theta+\tan\theta)$ $=\Big[\sec\theta-\frac{\sin\theta}{\cos\theta}\Big]\times\big(\sec\theta+\tan\theta\big)$ $=(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)$ $=\big(\sec^2\theta-\tan^2\theta\big)=1$ $=\text{R.H.S.}$$\therefore\ \text{L.H.S.}=\text{R.H.S}$
View full question & answer→Question 264 Marks
Prove the following identities:
$\frac{1+\cos\theta-\sin^2\theta}{\sin\theta(1+\cos\theta)}=\cot\theta$
Answer$\text{LHS}=\frac{1+\cos\theta-\sin^2\theta}{\sin\theta(1+\cos\theta)}$
$=\frac{1+\cos\theta-\big(1-\cos^2\theta\big)}{\sin\theta(1+\cos\theta)}$
$=\frac{1+\cos\theta-1+\cos^2\theta}{\sin\theta(1+\cos\theta)}$
$=\frac{\cos\theta(1+\cos\theta)}{\sin\theta(1+\cos\theta)}$
$=\frac{\cos\theta}{\sin\theta}$
$=\cot\theta$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 274 Marks
Prove the following identities:
$\frac{\cot^2\theta(\sec\theta-1)}{(1+\sin\theta)}+\frac{\sec^2\theta(\sin\theta-1)}{(1+\sec\theta)}=0$
Answer$\text{LHS}=\frac{\cot^2\theta(\sec\theta-1)}{(1+\sin\theta)}+\frac{\sec^2\theta(\sin\theta-1)}{(1+\sec\theta)}$
$=\frac{\cot^2\theta(\sec\theta-1)(1+\sec\theta)+\sec^2\theta(\sin\theta-1)(1+\sin\theta)}{(1+\sin\theta)(1+\sec\theta)}$
$=\frac{\cot^2\theta\big(\sec^2\theta-1\big)+\sec^2\theta\big(\sin^2\theta-1\big)}{(1+\sin\theta)(1+\sec\theta)}$
$=\frac{\cot^2\theta\tan^2\theta+\sec^2\theta\big(-\cos^2\theta\big)}{(1+\sin\theta)(1+\sec\theta)}$
$=\frac{\cot^2\theta\tan^2\theta-\sec^2\theta\cos^2\theta}{(1+\sin\theta)(1+\sec\theta)}$
$=\frac{\cot^2\theta\times\frac{1}{\cot^2\theta}-\sec^2\theta\times\frac{1}{\sec^2\theta}}{(1+\sin\theta)(1+\sec\theta)}$
$=\frac{1-1}{(1+\sin\theta)(1+\sec\theta)}$
$=0$
$\therefore\ \text{R.H.S.}=\text{L.H.S.}$
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Prove the following identities:
$\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}+\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=\frac{2}{\big(2\sin^2\theta-1\big)}$
Answer$\text{LHS}=\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}+\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}$
$=\frac{(\sin\theta-\cos\theta)^2+(\sin\theta+\cos\theta)^2}{(\sin\theta+\cos\theta)(\sin\theta-\cos\theta)}$
$=\frac{\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta+\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta}{\sin^2\theta-\cos^2\theta}$
$=\frac{1+1}{\sin^2\theta-\big(1-\sin^2\theta\big)}$
$=\frac{2}{\big(2\sin^2\theta-1\big)}$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
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