Trigonometrical Identities — Important Questions

Answer: (b) $1$.

By the fundamental Pythagorean identity, $\sin^2\theta+\cos^2\theta=1$ for every angle $\theta$.

Answer: (c) $1$.

Since $1+\tan^2\theta=\sec^2\theta$, rearranging gives $\sec^2\theta-\tan^2\theta=1$.

Answer: (c) $1$.

Using complementary angles, $\cos42^\circ=\cos(90^\circ-48^\circ)=\sin48^\circ$, so $\dfrac{\sin48^\circ}{\cos42^\circ}=\dfrac{\sin48^\circ}{\sin48^\circ}=1$.

Answer: (a) $2$.

Squaring, $\left(\tan\theta+\dfrac{1}{\tan\theta}\right)^2=\tan^2\theta+\dfrac{1}{\tan^2\theta}+2=4$, so $\tan^2\theta+\dfrac{1}{\tan^2\theta}=4-2=2$.

Answer: (a) $\cos^2\theta(1+\tan^2\theta)=\cos^2\theta\cdot\sec^2\theta=\cos^2\theta\cdot\dfrac{1}{\cos^2\theta}=1$. R correctly explains A.

$(1-\sin^2\theta)\sec^2\theta$

$=\cos^2\theta\cdot\sec^2\theta$ (since $1-\sin^2\theta=\cos^2\theta$)

$=\cos^2\theta\cdot\dfrac{1}{\cos^2\theta}=1=\text{RHS}.$

Hence proved.

Using complementary angles: $\cot55^\circ=\cot(90^\circ-35^\circ)=\tan35^\circ$ and $\sin62^\circ=\sin(90^\circ-28^\circ)=\cos28^\circ$.

$\dfrac{\tan35^\circ}{\cot55^\circ}+\dfrac{\cos28^\circ}{\sin62^\circ}=\dfrac{\tan35^\circ}{\tan35^\circ}+\dfrac{\cos28^\circ}{\cos28^\circ}=1+1=2.$

Taking the LCM on the LHS:

$\text{LHS}=\dfrac{\sin^2\theta+(1+\cos\theta)^2}{(1+\cos\theta)\sin\theta}$

$=\dfrac{\sin^2\theta+1+2\cos\theta+\cos^2\theta}{(1+\cos\theta)\sin\theta}$

Since $\sin^2\theta+\cos^2\theta=1$:

$=\dfrac{1+1+2\cos\theta}{(1+\cos\theta)\sin\theta}=\dfrac{2+2\cos\theta}{(1+\cos\theta)\sin\theta}$

$=\dfrac{2(1+\cos\theta)}{(1+\cos\theta)\sin\theta}=\dfrac{2}{\sin\theta}=2\csc\theta=\text{RHS}.$

Hence proved.

Multiply numerator and denominator inside the root by $(1+\sin\theta)$:

$\sqrt{\dfrac{1+\sin\theta}{1-\sin\theta}}=\sqrt{\dfrac{(1+\sin\theta)^2}{(1-\sin\theta)(1+\sin\theta)}}=\sqrt{\dfrac{(1+\sin\theta)^2}{1-\sin^2\theta}}$

$=\sqrt{\dfrac{(1+\sin\theta)^2}{\cos^2\theta}}=\dfrac{1+\sin\theta}{\cos\theta}$

$=\dfrac{1}{\cos\theta}+\dfrac{\sin\theta}{\cos\theta}=\sec\theta+\tan\theta=\text{RHS}.$

Hence proved.

Squaring the given relation:

$(\sin\theta+\cos\theta)^2=(\sqrt3)^2$

$\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=3$

$1+2\sin\theta\cos\theta=3\Rightarrow \sin\theta\cos\theta=1.$

Now

$\tan\theta+\cot\theta=\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\dfrac{1}{\sin\theta\cos\theta}=\dfrac{1}{1}=1.$

Hence proved.

Write $\tan A=\dfrac{\sin A}{\cos A}$ and $\cot A=\dfrac{\cos A}{\sin A}$.

First term:

$\dfrac{\cos A}{1-\dfrac{\sin A}{\cos A}}=\dfrac{\cos A}{\dfrac{\cos A-\sin A}{\cos A}}=\dfrac{\cos^2 A}{\cos A-\sin A}.$

Second term:

$\dfrac{\sin A}{1-\dfrac{\cos A}{\sin A}}=\dfrac{\sin A}{\dfrac{\sin A-\cos A}{\sin A}}=\dfrac{\sin^2 A}{\sin A-\cos A}=-\dfrac{\sin^2 A}{\cos A-\sin A}.$

Adding:

$\dfrac{\cos^2 A}{\cos A-\sin A}-\dfrac{\sin^2 A}{\cos A-\sin A}=\dfrac{\cos^2 A-\sin^2 A}{\cos A-\sin A}$

$=\dfrac{(\cos A-\sin A)(\cos A+\sin A)}{\cos A-\sin A}=\cos A+\sin A=\sin A+\cos A=\text{RHS}.$

Hence proved.

Expand both squares on the LHS.

$(\sin\theta+\csc\theta)^2=\sin^2\theta+2\sin\theta\csc\theta+\csc^2\theta=\sin^2\theta+2+\csc^2\theta$ (since $\sin\theta\csc\theta=1$).

$(\cos\theta+\sec\theta)^2=\cos^2\theta+2\cos\theta\sec\theta+\sec^2\theta=\cos^2\theta+2+\sec^2\theta$ (since $\cos\theta\sec\theta=1$).

Adding:

$\text{LHS}=(\sin^2\theta+\cos^2\theta)+4+\csc^2\theta+\sec^2\theta$

$=1+4+\csc^2\theta+\sec^2\theta=5+\csc^2\theta+\sec^2\theta.$

Now use $\csc^2\theta=1+\cot^2\theta$ and $\sec^2\theta=1+\tan^2\theta$:

$=5+(1+\cot^2\theta)+(1+\tan^2\theta)=7+\tan^2\theta+\cot^2\theta=\text{RHS}.$

Hence proved.

(i) By the identity $\sin^2\theta+\cos^2\theta=1$ (here $\theta=30^\circ$), $\sin^2 30^\circ+\cos^2 30^\circ=1.$

(ii) $\tan45^\circ=1$, so $\tan45^\circ+\dfrac{1}{\tan45^\circ}=1+1=2.$

(iii) $\sin45^\circ=\dfrac{1}{\sqrt2}$ and $\cos60^\circ=\dfrac12$.

$2\left(\dfrac{1}{\sqrt2}\right)^2+3\left(\dfrac12\right)^2=2\cdot\dfrac12+3\cdot\dfrac14=1+\dfrac34=\dfrac74.$

(iv) $1-\cos^2 60^\circ=\sin^2 60^\circ$, so

$\dfrac{1-\cos^2 60^\circ}{\cos^2 60^\circ}=\dfrac{\sin^2 60^\circ}{\cos^2 60^\circ}=\tan^2 60^\circ.$

Hence shown. (Numerically both equal $3$.)