Question: Prove: $\dfrac{1 + sin \theta}{cos \theta} = \dfrac{1}{sec \theta - tan \theta}$. Solution: Taking RHS $= \dfrac{1}{sec \theta - tan \theta}$ $= \dfrac{1}{\frac{1}{cos \theta} - \frac{sin \theta}{cos \theta}}$ $= \dfrac{1}{\dfrac{1 - sin \theta}{cos \theta}}$ $= 1 × \dfrac{cos \theta}{1 - sin \theta}$ $= \dfrac{cos \theta}…

Question: Prove: $\dfrac{1}{1 - sinx} - \dfrac{1}{1 + sinx} = 2sinx.sec^2x$. Solution: Taking LHS $= \dfrac{1}{1 - sinx} - \dfrac{1}{1 + sinx}$ $= \dfrac{(1 + sinx) - (1 - sinx)}{(1-sinx)(1+ sinx)}$ $= \dfrac{1 + sinx -1 + sinx}{1 - sin^2x}$ [Using (1 - sin²x = cos²x)] $= \dfrac{2sinx}{cos^2 x}$ $= 2sinx × \dfrac{1}{cos^2x}$ [Using …

Question: Prove that: $\dfrac{tan x}{secx -1} - \dfrac{sinx}{1 + cosx} = 2cotx$. Solution: To prove :  $\dfrac{tan x}{secx -1} - \dfrac{sinx}{1 + cosx} = 2cotx$ Taking LHS $= \dfrac{tan x}{secx -1} - \dfrac{sinx}{1 + cosx}$ [$tanx = \frac{sinx}{cosx}$ and $secx= \frac{1}{cosx}$] $= \dfrac{ \frac{sinx}{cosx} }{ \frac{1}{cosx} -1} - \…

Question: Prove the following trigonometric identity: $1 + tan 4A.tan 2A = sec 4A$ Solution: Taking LHS $ = 1 + tan 4A.tan 2A$ $= 1 + tan(2×2A) . tan2A$ Remember: $tan2A = \dfrac{2 tanA}{1 - tan^2 A}$ $= 1 + \dfrac{2tan2A}{1 - tan^2 2A} . tan2A$ $= 1 + \dfrac{2 tan^2 2A}{1 - tan^2 2A}$ $= \dfrac{(1 - tan^2 2A) + 2tan^2 2A}{1 - tan^2…

Question - Prove that: $cos^6A + sin^6A =$ (i) $(1 - sin^2A cos^2A)$ (ii) $\left( 1 - \dfrac{sin^2 2A}{4} \right)$ (iii) $\dfrac{4 - sin^22A}{4}$ This is a class 10 Question From Multiple Angles chapter of Unit Trigonometry. Solution: Taking LHS $= cos^6 A + sin^6A$ $= (cos^2 A)^3 + (sin^2 A…

Question: Prove that $\dfrac{(1+tanA)² + (1 - tanA)²}{(1 + cotA)² + (1-cotA)²} = tan²A$ We know, (a + b)² = a² + 2ab + b² (a - b)² = a² - 2ab + b² So, (a + b)² + (a - b)² = a² + 2ab + b² + a² - 2ab + b² = 2(a² + b²) Solution: Taking LHS $= \dfrac{(1+tanA)² + (1 - tanA)²}{(1 + cotA)² …

Question: Prove that: $\dfrac{1 + cos2A}{2} = cos^2A$ Solution: First Method Using identity of 1 and formula cos2A = cos²A - sin²A Taking LHS $= \dfrac{1 + cos2A}{2}$ $= \dfrac{(cos^2 A + sin^2A) + (cos^2A - sin^2A}{2}$ $= \dfrac{2cos^2A}{2}$ $= cos^2A$ = RHS Second Method: Using formula cos2A = 2cos²A -1 Taking LHS …

Question: If $cos A$ = $\dfrac{1}{2} \left ( a +\dfrac{1}{a} \right)$, show that: $cos 2A = \dfrac{1}{2} \left ( a² + \dfrac{1}{a²} \right)$ Solution: Given, $cos A$ = $\dfrac{1}{2} \left ( a +\dfrac{1}{a} \right)$ We know, $cos 2A$ $= 2cos²A - 1$ $= 2 (cosA)² -1$ $= 2 \left \{  \dfrac{1}{2} \left ( a + \dfrac{1}{a} \right) \right \…

Question: Prove the following trigonometric identity: sin5A = 16 sin⁵A - 20sin³A +5 sinA Solution: LHS = sin 5A = sin (2A +3A) [  sin(A+B) = sinA.cosB + cosA.sinB] = sin 2A . cos 3A + cos 2A . sin 3A = (2sinA cosA).(4cos³A - 3cosA) + (1 -2sin²A).(3sinA - 4sin³A) = 8sinA cos⁴A - 6sinAcos²A + 3sinA - 4sin³A - 6sin³A +8sin⁵A = 8sin⁵A +…

Prove the following trigonometric identity: $cot (45° -A)$ = $tan 2A + sec 2A$ Solution: LHS = $cot (45° -A)$ = $\dfrac{cot45°.cotA + 1}{cotA - cot45°}$ = $\dfrac{cot A + 1}{cot A - 1}$ = $\dfrac{\frac{cosA}{sinA} +1}{\frac{cosA}{sinA} -1}$ = $\dfrac{\frac{cosA +sinA}{sinA}}{\frac{cosA -sinA}{sinA}}$ = $\dfrac{cosA +sinA}{cosA -sinA…

Question: Prove the following trigonometric identity: $cos⁴\dfrac{A}{2} - sin⁴\dfrac{A}{2}$ = cos A. Solution: LHS = $cos⁴\dfrac{A}{2} - sin⁴\dfrac{A}{2}$ = $(cos²\dfrac{A}{2})² - (sin²\dfrac{A}{2})²$ = $(cos²\dfrac{A}{2} - sin²\dfrac{A}{2}) * (cos²\dfrac{A}{2} + sin²\dfrac{A}{2})$ = $1 * cos(2*\dfrac{1}{2})$ = $cos A$ = RHS Related…

Question: Prove the following trigonometric identity: $\dfrac{sin 8A}{sin 4A} - \dfrac{cos 8A}{cos 4A}$ = sec 4A. Solution: LHS = $\dfrac{sin 8A}{sin 4A} - \dfrac{cos 8A}{cos 4A}$ = $\dfrac{sin 8A.cos4A - cos8A.sin4A}{sin 4A.cos 4A}$ = $\dfrac{sin (8A -4A)}{sin 4A.cos4A}$ = $\dfrac{sin 4A}{sin4A.cos4A}$ = $\dfrac{1}{cos 4A}$ = $sec …

Question: Prove the following trigonometric identity: $\dfrac{sin 2A}{sinA} - \dfrac{cos 2A}{cosA}$ = sec A. Solution: LHS = $\dfrac{sin 2A}{sinA} - \dfrac{cos 2A}{cosA}$ = $\dfrac{sin 2A.cosA - cos 2A.sinA}{sinA.cosA}$ = $\dfrac{sin (2A-A)}{sinA.cosA}$ = $\dfrac{sin A}{sinA.cosA}$ = $\dfrac{1}{cosA}$ = $sec A$ RHS Related Notes And…

Question: Prove the following trigonometric identity: $\dfrac{cosA}{cosA -sinA} - \dfrac{cosA}{cosA +sinA}$ = tan2A Solution: LHS = $\dfrac{cosA}{cosA -sinA} - \dfrac{cosA}{cosA +sinA}$ = $\dfrac{cosA(cosA +sinA) -cosA(cosA -sinA}{(cosA -sinA)(cosA +sinA)}$ = $\dfrac{cosA (cosA +sinA -cosA +sinA}{cos²A -sin²A}$ = $\dfrac{cosA(2sinA)…

Question: Prove the following trigonometric identity: tan2A +sin2A = $\dfrac{4tanA}{1-tan⁴A}$ Solution: Given, LHS = tan 2A + sin 2A = $\dfrac{2 tanA}{1 -tan²A} + \dfrac{2 tanA}{1 +tan²A}$ = $\dfrac{2 tanA(1 +tan²A) + 2tanA (1 -tan²A)}{(1 -tan²A)(1 +tan²A)}$ = $\dfrac{2tanA (1 +tan²A +1 -tan²A}{1 -tan⁴A}$ = $\dfrac{2tanA (2)}{1 -tan…

Question: Prove the following trigonometric identity: $\dfrac{sin5A}{sinA} - \dfrac{cos5A}{cosA}$ = 4 cos2A. Solution: Given, LHS = $\dfrac{sin 5A}{sinA} - \dfrac{cos 5A}{cosA}$ = $\dfrac{sin 5A.cosA - cos5A.sinA}{sinAcosA}$ = $\dfrac{sin(5A -A)}{sinAcosA}$ = $\dfrac{sin4A}{sinAcosA}$ = $\dfrac{sin(2*2A)}{sinAcosA}$ = $\dfrac{2sin2A…

Question: Prove the following trigonometric identity: $\dfrac{sin 3A}{sinA} - \dfrac{cos 3A}{cosA}$ = 2. Solution: You can also solve the above question by the following process using the formula of compound angles. Solution: Given, LHS = $\dfrac{sin 3A}{sinA} - \dfrac{cos 3A}{cosA}$ = $\dfrac{sin 3A.cosA - cos3A.sinA}{sinAcosA}$ …

If tanA = x/y, prove that y.cos2A +x.sin2A = y. This is a class 10 Question From Trigonometric Identities chapter of Unit Trigonometry. All the steps for the solutions are mentioned as hint. If that's hard for you to navigate, you can always visit the facebook link given at the end of every posts.  Solution : Given, tan A = $\…

Prove that : $\dfrac{1}{tan3A +tanA} -\dfrac{1}{cot3A +cotA} = cot 4A$ This is a class 10 Question From Trigonometric Identities chapter of Unit Trigonometry. All the steps for the solutions are mentioned as hint. If that's hard for you to navigate, you can always visit the facebook link given at the end of every posts.  Solutio…

Prove that : $\dfrac{1+sin2A -cos2A}{1+sin2A+cos2A} = tanA$ This is a class 10 Question From Trigonometric Identities chapter of Unit Trigonometry. All the steps for the solutions are mentioned as hint. If that's hard for you to navigate, you can always visit the facebook link given at the end of every posts.  Solution : Taking …