Frank solutions for Mathematics - Part 2 [English] Class 10 ICSE chapter 21 - Trigonometric Identities [Latest edition]

Prove the following identity :

${\displaystyle (1-{\sin }^2\theta ){\sec }^2\theta =1}$

Prove the following identity :

${\displaystyle (1-{\cos }^2\theta ){\sec }^2\theta ={\tan }^2\theta }$

Prove the following identity :

tanA+cotA=secAcosecA 

Prove the following identity :

${\displaystyle \sin \theta (1+\tan \theta )+\cos \theta (1+\cot \theta )=\sec \theta +\cos ec\theta }$ 

Prove the following identity :

 ( 1 + cotθ - cosecθ) ( 1 + tanθ + secθ) 

Prove the following identity :

sinθcotθ + sinθcosecθ = 1 + cosθ  

Prove the following identity :

secA(1 - sinA)(secA + tanA) = 1

Prove the following identity :

secA(1 + sinA)(secA - tanA) = 1

Prove the following identity :

cosecθ(1 + cosθ)(cosecθ - cotθ) = 1

Prove the following identity : 

${\displaystyle \frac{\sec A-1}{\sec A+1}=\frac{1-\cos A}{1+\cos A}}$

Prove the following identity :

${\displaystyle \frac{1+\sin A}{1-\sin A}=\frac{\cos ecA+1}{\cos ecA-1}}$

Prove the following identity :

${\displaystyle \frac{\cos A}{1+\sin A}=\sec A-\tan A}$

Prove the following identity :

${\displaystyle \frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}=\frac{1+\sin \theta }{\cos \theta }}$

Prove the following identity : 

${\displaystyle {\sin }^{2}A{\cos }^{2}B-{\cos }^{2}A{\sin }^{2}B={\sin }^{2}A-{\sin }^{2}B}$

Prove the following identity :

${\displaystyle {(1-\tan A)}^2+{(1+\tan A)}^2=2{\sec }^{2}A}$

Prove the following identity :

${\displaystyle \cos e{c}^{4}A-\cos e{c}^{2}A={\cot }^{4}A+{\cot }^{2}A}$

Prove the following identity :

${\displaystyle {\sec }^{2}A+\cos e{c}^{2}A={\sec }^{2}A\cos e{c}^{2}A}$

Prove the following identity :

${\displaystyle {\cos }^{4}A-{\sin }^{4}A=2{\cos }^{2}A-1}$

Prove the following identity :

${\displaystyle {\tan }^{2}A-{\sin }^{2}A={\tan }^{2}A.{\sin }^{2}A}$

Prove the following identity :

(secA - cosA)(secA + cosA) = ${\displaystyle {\sin }^{2}A+{\tan }^{2}A}$

Prove the following identity :

${\displaystyle {(\cos A+\sin A)}^2+{(\cos A-\sin A)}^2=2}$

Prove the following identity :

${\displaystyle (\cos ecA-\sin A)(\sec A-\cos A)(\tan A+\cot A)=1}$

Prove the following identity :

${\displaystyle {\sec }^{2}A.\cos e{c}^{2}A={\tan }^{2}A+{\cot }^{2}A+2}$

Prove the following identity : 

${\displaystyle \frac{\cot A+\cos ecA-1}{\cot A-\cos ecA+1}=\frac{\cos A+1}{\sin A}}$

Prove the following identity : 

${\displaystyle \frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\sin A+\cos A}$

Prove the following identity : 

${\displaystyle (\cos ecA-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}}$

Prove the following identity : 

${\displaystyle {\sin }^{4}A+{\cos }^{4}A=1-2{\sin }^{2}A{\cos }^{2}A}$

Prove the following Identities :

${\displaystyle \frac{\cos ecA}{\cot A+\tan A}=\cos A}$

Prove the following identities:

${\displaystyle \frac{\tan \text{A}+\tan \text{B}}{\cot \text{A}+\cot \text{B}}=\tan \text{A}\tan \text{B}}$

Prove the following identities:

${\displaystyle \frac{\sec \text{A}-1}{\sec \text{A}+1}={\left(\frac{\sin \text{A}}{1+\cos \text{A}}\right)}^2}$

Prove the following identity : 

${\displaystyle \frac{\sin A}{1+\cos A}+\frac{1+\cos A}{\sin A}=2\cos ecA}$

Prove the following identity :

${\displaystyle \frac{1+\cos A}{1-\cos A}={(\cos ecA+\cot A)}^2}$

Prove the following identity :

${\displaystyle \frac{\cot A+\tan B}{\cot B+\tan A}=\cot A\tan B}$

Prove the following identity :

${\displaystyle \frac{1}{\tan A+\cot A}=\sin A\cos A}$

Prove the following identity :

${\displaystyle \tan A-\cot A=\frac{1-2{\cos }^{2}A}{\sin A\cos A}}$

Prove the following identity : 

${\displaystyle \frac{(1+{\tan }^{2}A)\cot A}{\cos e{c}^{2}A}=\tan A}$

Prove the following identity : 

${\displaystyle \cos ecA+\cot A=\frac{1}{\cos ecA-\cot A}}$

Prove the following identity : 

${\displaystyle \frac{\cos ecA}{\cos ecA-1}+\frac{\cos ecA}{\cos ecA+1}=2{\sec }^{2}A}$

Prove the following identity : 

${\displaystyle \frac{1+\cos A}{1-\cos A}=\frac{{\tan }^{2}A}{{(\sec A-1)}^2}}$

Prove the following identity : 

${\displaystyle {(\cot A-\cos ecA)}^2=\frac{1-\cos A}{1+\cos A}}$

Prove the following identity : 

${\displaystyle \sqrt{\cos e{c}^{2}q-1}=\text{cosq cosecq}}$

Prove the following identity : 

${\displaystyle \sqrt{\frac{1+\sin q}{1-\sin q}}+\sqrt{\frac{1-\sin q}{1+\sin q}}}$ = 2secq

Prove the following identity : 

${\displaystyle \sqrt{\frac{1-\cos A}{1+\cos A}}=\frac{\sin A}{1+\cos A}}$

Prove the following identity : 

${\displaystyle \sqrt{\frac{1+\cos A}{1-\cos A}}=\cos ecA+\cot A}$

Prove the following identity : 

${\displaystyle \sqrt{\frac{\sec q-1}{\sec q+1}}+\sqrt{\frac{\sec q+1}{\sec q-1}}}$ = 2 cosesq

Prove the following identity : 

${\displaystyle {(\sec \theta -\tan \theta )}^2=\frac{1-\sin \theta }{1+\sin \theta }}$

Prove the following identity : 

${\displaystyle \frac{1}{\sin A+\cos A}+\frac{1}{\sin A-\cos A}=\frac{2\sin A}{1-2{\cos }^{2}A}}$

Prove the following identity : 

${\displaystyle \frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{2{\sin }^{2}A-1}}$

Prove the following identity : 

${\displaystyle {\tan }^{2}A-{\tan }^{2}B=\frac{{\sin }^{2}A-{\sin }^{2}B}{{\cos }^{2}A{\cos }^{2}B}}$

Prove the following identity : 

${\displaystyle \frac{\cos A}{1-\tan A}+\frac{{\sin }^{2}A}{\sin A-\cos A}=\cos A+\sin A}$

Prove the following identity : 

${\displaystyle (1+{\tan }^{2}A)+(1+\frac{1}{{\tan }^{2}A})=\frac{1}{{\sin }^{2}A-{\sin }^{4}A}}$

Prove the following identity : 

${\displaystyle \frac{{\cos }^{3}A+{\sin }^{3}A}{\cos A+\sin A}+\frac{{\cos }^{3}A-{\sin }^{3}A}{\cos A-\sin A}=2}$

Prove the following identity : 

${\displaystyle {(\tan \theta +\frac{1}{\cos \theta })}^2+{(\tan \theta -\frac{1}{\cos \theta })}^2=2\left(\frac{1+{\sin }^2\theta }{1-{\sin }^2\theta }\right)}$

Prove the following identity : 

${\displaystyle \frac{\sin A-\sin B}{\cos A+\cos B}+\frac{\cos A-\cos B}{\sin A+\sin B}=0}$

Prove the following identity : 

${\displaystyle \frac{1}{\cos A+\sin A-1}+\frac{2}{\cos A+\sin A+1}=\cos ecA+\sec A}$

Prove the following identity : 

${\displaystyle \frac{\cot A+\cos ecA-1}{\cot A-\cos ecA+1}=\frac{\cos A+1}{\sin A}}$

Prove the following identity :

${\displaystyle \frac{\sec A-1}{\sec A+1}=\frac{{\sin }^{2}A}{{(1+\cos A)}^2}}$

Prove the following identity  :

${\displaystyle {(1+\cot A)}^2+{(1-\cot A)}^2=2\cos e{c}^{2}A}$

Prove the following identity : 

${\displaystyle \frac{\cos ec\theta }{\tan \theta +\cot \theta }=\cos \theta }$

Prove the following identity : 

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

Prove the following identity : 

${\displaystyle \frac{1+\sin \theta }{\cos ec\theta -\cot \theta }-\frac{1-\sin \theta }{\cos ec\theta +\cot \theta }=2(1+\cot \theta )}$

Prove the following identity : 

${\displaystyle (1+\cot A+\tan A)(\sin A-\cos A)=\frac{\sec A}{\cos e{c}^{2}A}-\frac{\cos ecA}{{\sec }^{2}A}}$

Prove the following identity : 

${\displaystyle 2({\sin }^6\theta +{\cos }^6\theta )-3({\sin }^4\theta +{\cos }^4\theta )+1=0}$

Prove the following identity : 

${\displaystyle {\sin }^8\theta -{\cos }^8\theta =({\sin }^2\theta -{\cos }^2\theta )(1-2{\sin }^2\theta {\cos }^2\theta )}$

Prove the following identity : 

${\displaystyle {\sec }^{4}A-{\sec }^{2}A=\frac{{\sin }^{2}A}{{\cos }^{4}A}}$

Prove the following identity :

${\displaystyle \frac{{\tan }^2\theta }{{\tan }^2\theta -1}+\frac{\cos e{c}^2\theta }{{\sec }^2\theta -\cos e{c}^2\theta }=\frac{1}{{\sin }^2\theta -{\cos }^2\theta }}$

Prove the following identity :

${\displaystyle \frac{{\sec }^2\theta -{\sin }^2\theta }{{\tan }^2\theta }=\cos e{c}^2\theta -{\cos }^2\theta }$

Prove the following identity :

${\displaystyle \frac{{\cos }^3\theta +{\sin }^3\theta }{\cos \theta +\sin \theta }+\frac{{\cos }^3\theta -{\sin }^3\theta }{\cos \theta -\sin \theta }=2}$

Prove the following identity :

${\displaystyle \frac{\tan \theta +\sin \theta }{\tan \theta -\sin \theta }=\frac{\sec \theta +1}{\sec \theta -1}}$

Prove the following identity : 

${\displaystyle [\frac{1}{({\sec }^2\theta -{\cos }^2\theta )}+\frac{1}{(\cos e{c}^2\theta -{\sin }^2\theta )}]\left({\sin }^2\theta {\cos }^2\theta \right)=\frac{1-{\sin }^2\theta {\cos }^2\theta }{2+{\sin }^2\theta {\cos }^2\theta }}$

Prove the following identity :

${\displaystyle \frac{{\cot }^2\theta (\sec \theta -1)}{(1+\sin \theta )}={\sec }^2\theta \left(\frac{1-\sin \theta }{1+\sec \theta }\right)}$

If m = a secA + b tanA and n = a tanA + b secA , prove that m² - n² = a² - b²

If ${\displaystyle \frac{x}{a\cos \theta }=\frac{y}{b\sin \theta } \text{and} \frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2,\text{prove that} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$

If x = r sinA cosB , y = r sinA sinB and z = r cosA , prove that   ${\displaystyle x^2+y^2+z^2=r^2}$

If sinA + cosA = m and secA + cosecA = n , prove that n(m² - 1) = 2m

If x = acosθ , y = bcotθ , prove that ${\displaystyle \frac{a^2}{x^2}-\frac{b^2}{y^2}=1.}$

If secθ + tanθ = m , secθ - tanθ = n , prove that mn = 1

If x = asecθ + btanθ and y = atanθ + bsecθ , prove that ${\displaystyle x^2-y^2=a^2-b^2}$

If tanA + sinA = m and tanA - sinA = n , prove that (${\displaystyle m^2-n^2{)}^2}$ = 16mn 

If sinA + cosA = ${\displaystyle \sqrt{2}}$ , prove that sinAcosA = ${\displaystyle \frac{1}{2}}$

If ${\displaystyle a{\sin }^2\theta +b{\cos }^2\theta =c\text{and}p{\sin }^2\theta +q{\cos }^2\theta =r}$ , prove that (b - c)(r - p) = (c - a)(q - r)

Without using trigonometric table , evaluate : 

${\displaystyle \cos ec49°\cos 41°+\frac{\tan 31°}{\cot 59°}}$

Without using trigonometric table , evaluate : 

${\displaystyle {\left(\frac{{\sin 47}^{\circ }}{{\cos 43}^{\circ }}\right)}^2-4{\cos }^2{45}^{\circ }+{\left(\frac{{\cos 43}^{\circ }}{{\sin 47}^{\circ }}\right)}^2}$

Without using trigonometric table , evaluate : 

${\displaystyle {\cos 90}^{\circ }+{\sin 30}^{\circ }{\tan 45}^{\circ }{\cos }^2{45}^{\circ }}$

Without using trigonometric table , evaluate : 

${\displaystyle {\left(\frac{{\sin 49}^{\circ }}{{\sin 41}^{\circ }}\right)}^2+{\left(\frac{{\cos 41}^{\circ }}{{\sin 49}^{\circ }}\right)}^2}$

Without using trigonometric table , evaluate : 

${\displaystyle \frac{{\sin 72}^{\circ }}{{\cos 18}^{\circ } }-\frac{{\sec 32}^{\circ }}{\cos ec{58}^{\circ }}}$

Find the value of ${\displaystyle \theta (0^{\circ }<\theta <{90}^{\circ })}$ if : 

${\displaystyle {\cos 63}^{\circ }\sec ({90}^{\circ }-\theta )=1}$

Find the value of ${\displaystyle \theta (0^{\circ }<\theta <{90}^{\circ })}$ if : 

${\displaystyle {\tan 35}^{\circ }\cot ({90}^{\circ }-\theta )=1}$

Without using trigonometric identity , show that :

${\displaystyle {\sin 42}^{\circ }{\sec 48}^{\circ }+{\cos 42}^{\circ }\cos ec{48}^{\circ }=2}$

Without using trigonometric identity , show that :

${\displaystyle {\tan 10}^{\circ }{\tan 20}^{\circ }{\tan 30}^{\circ }{\tan 70}^{\circ }{\tan 80}^{\circ }=\frac{1}{\sqrt{3}}}$

Without using trigonometric identity , show that :

${\displaystyle \sin ({50}^{\circ }+\theta )-\cos ({40}^{\circ }-\theta )=0}$

Without using trigonometric identity , show that :

${\displaystyle {\cos }^2{25}^{\circ }+{\cos }^2{65}^{\circ }=1}$

Without using trigonometric identity , show that :

${\displaystyle {\sec 70}^{\circ }{\sin 20}^{\circ }-{\cos 20}^{\circ }\cos ec{70}^{\circ }=0}$

Prove that ${\displaystyle \frac{\sin A}{\sin ({90}^{\circ }-A)}+\frac{\cos A}{\cos ({90}^{\circ }-A)}=\sec ({90}^{\circ }-A)\cos ec({90}^{\circ }-A)}$

For ΔABC , prove that : 

${\displaystyle \tan \left(\frac{B+C}{2}\right)=\cot \frac{A}{2}}$

For ΔABC , prove that : 

${\displaystyle \sin \left(\frac{A+B}{2}\right)=\cos \frac{C}{2}}$

Prove that  ${\displaystyle \sin ({90}^{\circ }-A).\cos ({90}^{\circ }-A)=\frac{\tan A}{1+{\tan }^{2}A}}$

Find the value of x , if ${\displaystyle \cos x={\cos 60}^{\circ }{\cos 30}^{\circ }-{\sin 60}^{\circ }{\sin 30}^{\circ }}$

Find x , if ${\displaystyle \cos (2x-6)={\cos }^2{30}^{\circ }-{\cos }^2{60}^{\circ }}$

Given ${\displaystyle {\cos 38}^{\circ }\sec ({90}^{\circ }-2A)=1}$ , Find the value of <A

prove that ${\displaystyle \frac{1}{1+\cos ({90}^{\circ }-A)}+\frac{1}{1-\cos ({90}^{\circ }-A)}=2\cos e{c}^2({90}^{\circ }-A)}$