Trigonometry - 10
Hello friend !! Aaj hum aapko trigonometric ratio of 45 degree ke baare me full details me sikhane waale hai. Aaj hum aapko trigonometry ke sabhi ratios ke 45 degree ke maan jaise - sin 45°, cos 45°, tan 45°, cosec 45°, sec 45° aur cot 45° ki value (maan) nikalana sikhayenge.
Friend agar aap humare new friend hai toh hum aapko bata de ki, Aaj ka lesson, trigonometry topic ka part-10 hai. Ab tak humne trigonometry ke 9 part likh kar aapko sikha chuke hai. Agar aap wo sabhi part padhana chahte hai toh niche humne sabhi part ka ek link diya hai. Aap niche diye link se sabhi parts padh kar sikh sakte hai.
❒ Trigonometry sabhi part yaha padhe.
Toh ab hum apne topic par aate hai. Aapne jarur padha hoga ki, sin45° = 1/√2, cos45° = 1/√2, tan45° = 1... cot45° = 1 hoti hai. Lekin kya aapne kabhi socha ki, inn sabhi ki value nikalte kaise hai? Matlab, hum trigonometry ratios sin 45°, cos 45°, tan 45°, coses 45°, sec 45°, aur cot 45° ki value (maan) kaise nikalte hai?
Aaj hum aapko yahi sikhane waale hai. Aaj hum aapko bilkul basic se issko sikhayenge.
Aap ekdam dhyaan se issko sikhe aur agar kahi koi bhi problem aaye toh turant comment karke jarur puchiye. Chaliye toh aajka lesson start karte hai.
∴ Right angled triangle ∆ABC se hum paate hai ki,
⇒ sin 45° = BC/AC
= a/a√2
= 1/√2
∴ sin 45° = 1/√2
⇒ cos 45° = AB/AC
= a/a√2
= 1/√2
∴ cos 45° = 1/√2
⇒ tan 45° = BC/AB
= a/a
= 1
∴ tan45° = 1
⇒ cosec 45° = AC/BC
= a√2/a
= √2
∴ cosec 45° = √2
⇒ sec 45° = AC/AB
= a√2/a
= √2
∴ sec 45° = √2
Que-1. Yadi sin 45° = 1/√2 ho toh cosec 45° ka maan kya hoga.
Sol.- Diya hai ki, sin 45° = 1/√2, toh Reciprocal relations se,
⇒ cosec 45° = 1/sin 45°
= 1/(1/√2)
= √2
∴ cosec 45° = √2 Ans.
Que-2. cos²45°+tan²45° ka maan batayiye.
Sol.- Diya hai ki, cos²45°+tan²45° ka maan batayiye.
Toh cos45° = 1/√2
∴ cos²45° = (1/√2)²
= 1/2
Issi prakar se, tan45° = 1
∴ tan²45 = 1² = 1
Tab cos²45° aur tan²45° ka maan diye huye question me rakh kar solve karne par
⇒ cos²45°+tan²45° = (1/2)+1
= 3/2
∴ cos²45°+tan²45° = 3/2 Ans.
Que-3. Yadi cosec θ = √2 , toh θ ka maan kya hoga. (0°<θ<90°)
Sol.- Diya hai ki, cosec θ = √2
⇒ cosec θ = cosec 45°
⇒ θ = 45° Ans.
Que-4. sec45°cot45°+sin45°cos45° ka maan kya hoga?
Sol.- Hum jaante hai ki, sec45° = √2, cot45° = 1, sin45° = 1/√2 aur cos45° = 1/√2
Sabhi maan question me rakhne par
⇒ sec45°cot45°+sin45°cos45°
= √2×1+(1/√2)×(1/√2)
= √2+1/2
= (2√2+1)/2 Ans.
Que-5. sin45°cos45°+cos45°sin45° ka maan gyaat kijiye.
Sol.- Hum jaante hai ki, sin45° = 1/√2 aur cos45° = 1/√2 toh,
⇒ sin45°cos45°+cos45°sin45°
= (1/√2)×(1/√2)+(1/√2)×(1/√2)
= (1/2)+(1/2)
= (1+1)/2
= 2/2
= 1 Ans.
Que-6. Yadi sinθ = cosθ ho, toh "θ" ka maan gyaat kijiye.
Sol.- Diya hai ki, sinθ = cosθ
⇒ sinθ/cosθ = 1 .......(A)
Aur hum jaante hai ki, sinθ, cosθ aur tanθ ke relation se,
sinθ/cosθ = tanθ
Toh point (A) se, tanθ = 1
⇒ tanθ = tan45°
⇒ θ = 45° Ans.
Que-7. Yadi cos²30°+cos²45° = X, toh X ka gyaat kijiye.
Sol.- cos30° = √3/2 aur cos45° = 1/√2
⇒ cos²30°+cos²45° = X
⇒ (√3/2)²+(1/√2)² = X
⇒ (3/4)+(1/2) = X
⇒ (3+2)/4 = X
⇒ 5/4 = X
⇒ X = 5/4 Ans.
Que-8. Yadi secα = √2 aur cotβ = √3 ho, toh (α+β) ka maan gyaat kijiye.
Sol.- secα = √2 ⇒ secα = sec45°
⇒ α = 45°
Aur cotβ = √3 ⇒ cotβ = cot30°
⇒ β = 30°
∴ (α+β) = (45°+30°) = 75° Ans.
Que-9. Yadi tanα×cotα+sinα = 2 aur secβ×cosβ+cosecβ = √2 ho toh sin²(α-β) ka maan gyaat kijiye.
Sol.- Diya hai ki, tanα×cotα+sinα = 2
⇒ 1+sinα = 2
∵Trigonometric ratios ke reciprocal relations
⇒ sinα = 2-1
⇒ sinα = 1
⇒ sinα = sin90°
⇒ α = 90°
Aur secβ×cosβ+tanβ = 2
⇒ 1+tanβ = 2
⇒ tanβ = 2-1
⇒ tanβ = 1
⇒ tanβ = tan45°
⇒ β = 45°
Ab sin²(α-β) = sin²(90-45)
= sin²45°
= (1/√2)²
= 1/2 Ans.
Que-10. Yadi 2xsinθ = x ho toh, [cot(θ+15°)+sec(θ+15°)]² ka maan gyaat kijiye.
Sol.- Diya hai ki, 2xsinθ = x
⇒ sinθ = x/2x
⇒ sinθ = 1/2
Ye bhi padhe-:
❒ Reciprocal relations between trigonometric ratios
❒ Rational numbers Defination with Examples
❒ Quadratic polynomial ko binomial se fast and easily divide kaise karte hai ?
Hume ummid hai ki, aapko aaj ka part-10 achhe samjh aaya hoga. Agar aapko koi doubt, problem ya kuch aapko puchhana hai toh aap comment karke puchh sakte hai.
Aap chahe toh apna question likhkar ya question ki image humare Facebook page (jo niche diya hai) par massage karke bhi send kar sakte hai. Hum jarur aapke question ka reply karenge.
Agar aapko aaj lesson helpful laga ho toh please ek share jarur kariye. Aapke ek share karne se hume aapki help karne me bahut energy milati hai aur hum double energy ke saath aapki help karte hai toh please Share jarur kariye.
Hello friend !! Aaj hum aapko trigonometric ratio of 45 degree ke baare me full details me sikhane waale hai. Aaj hum aapko trigonometry ke sabhi ratios ke 45 degree ke maan jaise - sin 45°, cos 45°, tan 45°, cosec 45°, sec 45° aur cot 45° ki value (maan) nikalana sikhayenge.
Friend agar aap humare new friend hai toh hum aapko bata de ki, Aaj ka lesson, trigonometry topic ka part-10 hai. Ab tak humne trigonometry ke 9 part likh kar aapko sikha chuke hai. Agar aap wo sabhi part padhana chahte hai toh niche humne sabhi part ka ek link diya hai. Aap niche diye link se sabhi parts padh kar sikh sakte hai.
❒ Trigonometry sabhi part yaha padhe.
Toh ab hum apne topic par aate hai. Aapne jarur padha hoga ki, sin45° = 1/√2, cos45° = 1/√2, tan45° = 1... cot45° = 1 hoti hai. Lekin kya aapne kabhi socha ki, inn sabhi ki value nikalte kaise hai? Matlab, hum trigonometry ratios sin 45°, cos 45°, tan 45°, coses 45°, sec 45°, aur cot 45° ki value (maan) kaise nikalte hai?
Aaj hum aapko yahi sikhane waale hai. Aaj hum aapko bilkul basic se issko sikhayenge.
Aap ekdam dhyaan se issko sikhe aur agar kahi koi bhi problem aaye toh turant comment karke jarur puchiye. Chaliye toh aajka lesson start karte hai.
Trigonometric ratios of 45 degree
Trigonometric ratios 45 degree means "45° कोण के लिए त्रिकोणमिति अनुपात " nikalane ke liye hum sab se pahle ek right angled triangle lenge. Right angled triangle means "samkon tribhuj" (समकोण त्रिभुज). Samkon tribhuj, aisa triangle hota hai jisme kam-se-kam ek angle 90° ka jarur hota hai. Agar aap triangle ke baare me details me jaanana chahte hai toh humne nichey diya hai. Aap issko jarur padhe.
❒ Triangle and its types and defination in hindi
Maana nichey diya gaya ∆ABC ek right angled triangle hai. Jiska ∠B = 90° aur ∠A = ∠C = 45° hai.
Maana sides AB = BC = a hai toh Pythagoras theorem se,
AC² = AB²+BC²
= a²+a²
= 2a²
AC = a√2
❒ Triangle and its types and defination in hindi
Maana nichey diya gaya ∆ABC ek right angled triangle hai. Jiska ∠B = 90° aur ∠A = ∠C = 45° hai.
Maana sides AB = BC = a hai toh Pythagoras theorem se,
AC² = AB²+BC²
= a²+a²
= 2a²
AC = a√2
∴ Right angled triangle ∆ABC se hum paate hai ki,
⇒ sin 45° = BC/AC
= a/a√2
= 1/√2
∴ sin 45° = 1/√2
⇒ cos 45° = AB/AC
= a/a√2
= 1/√2
∴ cos 45° = 1/√2
⇒ tan 45° = BC/AB
= a/a
= 1
∴ tan45° = 1
⇒ cosec 45° = AC/BC
= a√2/a
= √2
∴ cosec 45° = √2
⇒ sec 45° = AC/AB
= a√2/a
= √2
∴ sec 45° = √2
⇒ cot 45° = AB/BC
= a/a
= 1
∴ cot 45° = 1
= a/a
= 1
∴ cot 45° = 1
Trigonometric Ratio 45 degree se Related Question and Solution
Ab hum aapko 45 degree se related kuch question ko solve karna sikhate hai. Aap inn questions ke solutions ko ekdam dhyaan se samjhiye. Agar kisi bhi point par kuch puchhana ho toh comment karke jarur puchiye.Que-1. Yadi sin 45° = 1/√2 ho toh cosec 45° ka maan kya hoga.
Sol.- Diya hai ki, sin 45° = 1/√2, toh Reciprocal relations se,
⇒ cosec 45° = 1/sin 45°
= 1/(1/√2)
= √2
∴ cosec 45° = √2 Ans.
Que-2. cos²45°+tan²45° ka maan batayiye.
Sol.- Diya hai ki, cos²45°+tan²45° ka maan batayiye.
Toh cos45° = 1/√2
∴ cos²45° = (1/√2)²
= 1/2
Issi prakar se, tan45° = 1
∴ tan²45 = 1² = 1
Tab cos²45° aur tan²45° ka maan diye huye question me rakh kar solve karne par
⇒ cos²45°+tan²45° = (1/2)+1
= 3/2
∴ cos²45°+tan²45° = 3/2 Ans.
Que-3. Yadi cosec θ = √2 , toh θ ka maan kya hoga. (0°<θ<90°)
Sol.- Diya hai ki, cosec θ = √2
⇒ cosec θ = cosec 45°
⇒ θ = 45° Ans.
Que-4. sec45°cot45°+sin45°cos45° ka maan kya hoga?
Sol.- Hum jaante hai ki, sec45° = √2, cot45° = 1, sin45° = 1/√2 aur cos45° = 1/√2
Sabhi maan question me rakhne par
⇒ sec45°cot45°+sin45°cos45°
= √2×1+(1/√2)×(1/√2)
= √2+1/2
= (2√2+1)/2 Ans.
Que-5. sin45°cos45°+cos45°sin45° ka maan gyaat kijiye.
Sol.- Hum jaante hai ki, sin45° = 1/√2 aur cos45° = 1/√2 toh,
⇒ sin45°cos45°+cos45°sin45°
= (1/√2)×(1/√2)+(1/√2)×(1/√2)
= (1/2)+(1/2)
= (1+1)/2
= 2/2
= 1 Ans.
Que-6. Yadi sinθ = cosθ ho, toh "θ" ka maan gyaat kijiye.
Sol.- Diya hai ki, sinθ = cosθ
⇒ sinθ/cosθ = 1 .......(A)
Aur hum jaante hai ki, sinθ, cosθ aur tanθ ke relation se,
sinθ/cosθ = tanθ
Toh point (A) se, tanθ = 1
⇒ tanθ = tan45°
⇒ θ = 45° Ans.
Que-7. Yadi cos²30°+cos²45° = X, toh X ka gyaat kijiye.
Sol.- cos30° = √3/2 aur cos45° = 1/√2
⇒ cos²30°+cos²45° = X
⇒ (√3/2)²+(1/√2)² = X
⇒ (3/4)+(1/2) = X
⇒ (3+2)/4 = X
⇒ 5/4 = X
⇒ X = 5/4 Ans.
Que-8. Yadi secα = √2 aur cotβ = √3 ho, toh (α+β) ka maan gyaat kijiye.
Sol.- secα = √2 ⇒ secα = sec45°
⇒ α = 45°
Aur cotβ = √3 ⇒ cotβ = cot30°
⇒ β = 30°
∴ (α+β) = (45°+30°) = 75° Ans.
Que-9. Yadi tanα×cotα+sinα = 2 aur secβ×cosβ+cosecβ = √2 ho toh sin²(α-β) ka maan gyaat kijiye.
Sol.- Diya hai ki, tanα×cotα+sinα = 2
⇒ 1+sinα = 2
∵Trigonometric ratios ke reciprocal relations
⇒ sinα = 2-1
⇒ sinα = 1
⇒ sinα = sin90°
⇒ α = 90°
Aur secβ×cosβ+tanβ = 2
⇒ 1+tanβ = 2
⇒ tanβ = 2-1
⇒ tanβ = 1
⇒ tanβ = tan45°
⇒ β = 45°
Ab sin²(α-β) = sin²(90-45)
= sin²45°
= (1/√2)²
= 1/2 Ans.
Que-10. Yadi 2xsinθ = x ho toh, [cot(θ+15°)+sec(θ+15°)]² ka maan gyaat kijiye.
Sol.- Diya hai ki, 2xsinθ = x
⇒ sinθ = x/2x
⇒ sinθ = 1/2
⇒ sinθ = sin30°
⇒ θ = 30°
Ab [cot(θ+15°)+sec(θ+15°)]²
= [cot(30°+15°)+sec(30°+15°)]²
= [cot45°+sec45°]²
= [1+√2]²
= (1)²+(√2)²+2×1×√2
= 1+2+2√2
= 3+2√2 Ans.
❒ Reciprocal relations between trigonometric ratios
❒ Rational numbers Defination with Examples
❒ Quadratic polynomial ko binomial se fast and easily divide kaise karte hai ?
Hume ummid hai ki, aapko aaj ka part-10 achhe samjh aaya hoga. Agar aapko koi doubt, problem ya kuch aapko puchhana hai toh aap comment karke puchh sakte hai.
Aap chahe toh apna question likhkar ya question ki image humare Facebook page (jo niche diya hai) par massage karke bhi send kar sakte hai. Hum jarur aapke question ka reply karenge.
Agar aapko aaj lesson helpful laga ho toh please ek share jarur kariye. Aapke ek share karne se hume aapki help karne me bahut energy milati hai aur hum double energy ke saath aapki help karte hai toh please Share jarur kariye.
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