Trigonometrical Identities — ICSE Class 10 Maths Important Questions

Answer: (b) $1$1.

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

Answer: (c) $1$1.

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

Answer: (c) $1$1.

Using complementary angles, $\cos42^\circ=\cos(90^\circ-48^\circ)=\sin48^\circ$42^=(90^-48^)=48^, so $\dfrac{\sin48^\circ}{\cos42^\circ}=\dfrac{\sin48^\circ}{\sin48^\circ}=1$48^/42^=48^/48^=1.

Answer: (a) $2$2.

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

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

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

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

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

Hence proved.

Using complementary angles: $\cot55^\circ=\cot(90^\circ-35^\circ)=\tan35^\circ$55^=(90^-35^)=35^ and $\sin62^\circ=\sin(90^\circ-28^\circ)=\cos28^\circ$62^=(90^-28^)=28^.

$\dfrac{\tan35^\circ}{\cot55^\circ}+\dfrac{\cos28^\circ}{\sin62^\circ}=\dfrac{\tan35^\circ}{\tan35^\circ}+\dfrac{\cos28^\circ}{\cos28^\circ}=1+1=2.$35^/55^+28^/62^=35^/35^+28^/28^=1+1=2.

Taking the LCM on the LHS:

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

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

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

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

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

Hence proved.

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

$\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}}$1+/1-=(1+)^2/(1-)(1+)=(1+)^2/1-^2

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

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

Hence proved.

Squaring the given relation:

$(\sin\theta+\cos\theta)^2=(\sqrt3)^2$(+)^2=(3)^2

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

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

Now

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

Hence proved.

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

First term:

$\dfrac{\cos A}{1-\dfrac{\sin A}{\cos A}}=\dfrac{\cos A}{\dfrac{\cos A-\sin A}{\cos A}}=\dfrac{\cos^2 A}{\cos A-\sin A}.$A1- A/ A= A A- A/ A=^2 A/ A- A.

Second term:

$\dfrac{\sin A}{1-\dfrac{\cos A}{\sin A}}=\dfrac{\sin A}{\dfrac{\sin A-\cos A}{\sin A}}=\dfrac{\sin^2 A}{\sin A-\cos A}=-\dfrac{\sin^2 A}{\cos A-\sin A}.$A1- A/ A= A A- A/ A=^2 A/ A- A=-^2 A/ A- A.

Adding:

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

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

Hence proved.

Expand both squares on the LHS.

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

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

Adding:

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

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

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

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

Hence proved.

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

(ii) $\tan45^\circ=1$45^=1, so $\tan45^\circ+\dfrac{1}{\tan45^\circ}=1+1=2.$45^+1/45^=1+1=2.

(iii) $\sin45^\circ=\dfrac{1}{\sqrt2}$45^=1/2 and $\cos60^\circ=\dfrac12$60^=12.

$2\left(\dfrac{1}{\sqrt2}\right)^2+3\left(\dfrac12\right)^2=2\cdot\dfrac12+3\cdot\dfrac14=1+\dfrac34=\dfrac74.$2(1/2)^2+3(12)^2=2·12+3·14=1+34=74.

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

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

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